Q.Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is
Concept understanding — Refraction at Spherical Surface
Refraction at a Spherical Surface
Imagine you're looking at a fish in a pond. The fish appears closer to the surface than it actually is. That's refraction — light bends when it moves from water to air. Now take that idea and replace the flat water surface with a curved one, like a glass lens or a drop of water. That's refraction at a spherical surface.
The core intuition
When light hits a flat surface (like a glass slab), it bends once and travels straight. But when the surface is curved — part of a sphere — the angle at which light hits changes depending on where on the surface it strikes. A ray hitting near the centre meets the surface almost head-on; a ray hitting near the edge meets it at a steep slant. This variation in incidence angle is what makes spherical surfaces focus or diverge light.
Think of a spherical surface as a tiny piece of a sphere. The centre of that sphere is called the centre of curvature (C). The distance from the surface to C is the radius of curvature (R). The line joining the centre of the surface (the pole, P) to C is the principal axis.
The precise geometry
We need to track what happens to a ray from an object point O on the principal axis. The ray travels in medium 1 (refractive index n1), hits the spherical surface at point A, and enters medium 2 (refractive index n2). The surface has radius R, with centre C.
The key is Snell's law at point A:
n1sini=n2sinr
But i and r are measured from the normal at A. For a spherical surface, the normal at any point is the line joining that point to C. So the normal is AC.
For small angles (paraxial rays — rays close to the axis), sinθ≈θ in radians. This approximation is the backbone of all standard lens and mirror formulas. It lets us replace Snell's law with:
n1i=n2r
Now look at the geometry. Let the object distance from the pole be u (negative by sign convention — object on left), and the image distance be v (positive if image is on the right, in medium 2). The angle the incident ray makes with the axis is α, the refracted ray makes β, and the normal makes θ with the axis.
From the triangles:
- In △OAC: i=α+θ
- In △AIC: r=θ−β (for a convex surface towards the object)
Substitute into Snell's law:
n1(α+θ)=n2(θ−β)
For small angles, α≈POAP≈−uh (since u is negative), β≈vh, and θ≈Rh.
Plugging these in:
n1(−uh+Rh)=n2(Rh−vh)
Cancel h (non-zero) and rearrange:
vn2−un1=Rn2−n1
This is the refraction at a spherical surface formula. It relates object distance u, image distance v, radii R, and the two refractive indices.
Sign convention (crucial for exams)
Use the Cartesian sign convention (the one used in NCERT and most Indian boards):
- Distances measured from the pole P along the principal axis.
- Positive in the direction of incident light (usually left to right).
- Negative opposite to incident light.
- R is positive if the centre of curvature C is on the right (convex surface towards object), negative if C is on the left (concave surface towards object).
The most common mistake is getting the sign of R wrong. Always check: is the centre of curvature on the same side as the incoming light or the opposite side? If opposite, R is positive.
What the formula tells you
- If n2>n1 (going from rarer to denser), the right side Rn2−n1 is positive for a convex surface. This means v is positive — the image forms on the other side (real image).
- If n2<n1 (denser to rarer), the right side is negative for a convex surface. The image may be virtual (on the same side as the object).
- If the surface is flat (R→∞), the formula reduces to vn2=un1, which is just Snell's law for a plane surface — the apparent depth formula.
A quick example
A small object is placed 30 cm in front of a convex spherical surface of radius 20 cm, separating air (n=1) from glass (n=1.5). Where is the image?
Here u=−30 cm, R=+20 cm (centre on the right), n1=1, n2=1.5.
v1.5−−301=201.5−1
v1.5+301=200.5=401
v1.5=401−301=1203−4=−1201
v=−180 cm
The negative v means the image is on the same side as the object — a virtual image 180 cm from the surface. This makes sense: a single convex surface between air and glass acts like a diverging lens for objects in air.
Why this matters
This single formula is the foundation for everything that follows: lenses (two spherical surfaces back-to-back), lens maker's formula, and even the human eye. Master this, and you've unlocked the geometry of how light bends at curved boundaries.
Refraction at a single spherical surface, n₂/v − n₁/u = (n₂−n₁)/R, is a foundational derivation in the NCERT Class 12 Physics chapter on ray optics, tested in CBSE boards, JEE Main and NEET as the basis for the lens maker's formula. Searches for "refraction at spherical surface formula derivation class 12 physics" will find this sign-convention-based approach matches the NCERT textbook exactly.
Why this formula?
Great — let’s build the Refraction at a Spherical Surface formula from first principles. The goal is to understand why the relation
vn2−un1=Rn2−n1
holds, where:
- n1 = refractive index of the first medium (where the object lies)
- n2 = refractive index of the second medium (where the image lies)
- u = object distance from the pole (sign convention: negative for real object)
- v = image distance from the pole (sign convention: positive for real image on the opposite side)
- R = radius of curvature of the spherical surface (positive if centre of curvature is on the image side)
1. The core idea: Snell’s law at a curved interface
At any point on the spherical surface, the incident ray and refracted ray obey Snell’s law:
n1sini=n2sinr
For small angles (paraxial approximation — rays close to the principal axis), sinθ≈θ (in radians). So:
n1i=n2r
This linearisation is the key that lets us turn geometry into algebra.
2. Geometry of a single ray
Consider a point object O on the principal axis. A ray from O strikes the spherical surface at point P (height h above the axis). Let:
- C = centre of curvature of the spherical surface
- M = pole of the surface (vertex)
- I = image point formed after refraction
Draw the normal at P — it passes through C (since the surface is spherical). The angles:
- i = angle between incident ray OP and the normal PC
- r = angle between refracted ray PI and the normal PC
3. Relating angles to distances (paraxial approximation)
Because h is small compared to u, v, and R:
- Angle between OP and the axis: α≈uh (with sign)
- Angle between PC (normal) and the axis: θ≈Rh
- Angle between PI and the axis: β≈vh
Now, from the geometry of the triangle formed by the ray, the normal, and the axis:
- Incident angle i = angle between OP and the normal = θ−α (if θ>α)
- Refracted angle r = angle between PI and the normal = θ−β
Check the sign convention carefully — the exact relation depends on whether the ray bends toward or away from the normal. For a convex surface (centre on the image side), the standard result is:
i=α+θandr=θ−β
But the difference that matters is:
i−r=α+β
4. Applying Snell’s law
From n1i=n2r, we can write:
n1i=n2(i−(i−r))or directly:
n1i=n2r⟹n1i−n2r=0
But it’s more useful to express r in terms of i and the geometry:
r=i−(α+β)
Substitute into Snell’s law:
n1i=n2[i−(α+β)]
Simplify:
n1i=n2i−n2(α+β)
(n1−n2)i=−n2(α+β)
Now, i≈α+θ (from geometry). For small angles, α≈h/u, β≈h/v, θ≈h/R.
5. Substituting the small-angle approximations
Let’s do it step by step:
(n1−n2)(α+θ)=−n2(α+β)
Replace α, β, θ:
(n1−n2)(uh+Rh)=−n2(uh+vh)
Cancel h (non-zero):
(n1−n2)(u1+R1)=−n2(u1+v1)
6. Rearranging to the standard form
Expand the left side:
un1−n2+Rn1−n2=−un2−vn2
Bring terms with 1/u together:
un1−n2+un2=−vn2−Rn1−n2
The left side simplifies:
un1−n2+n2=un1
So:
un1=−vn2−Rn1−n2
Multiply both sides by −1:
−un1=vn2+Rn1−n2
Finally, bring the 1/R term to the left:
vn2−un1=Rn2−n1
7. Why this formula makes physical sense
- If R→∞ (plane surface): the formula becomes vn2=un1, which is the familiar apparent depth formula for a plane interface.
- If n1=n2 (no refraction): the formula gives v1=u1, meaning v=u — the image coincides with the object (no bending).
- Sign of R determines whether the surface is convex or concave toward the incident ray — this flips the bending direction.
8. Key takeaway for exams
The derivation rests on three pillars:
- Snell’s law in the small-angle approximation: n1i=n2r
- Geometry of a circle: the normal at any point passes through the centre of curvature
- Paraxial approximation: tanθ≈θ≈distanceh
Memorise the final formula, but always recall that it comes from equating the bending of the ray (via Snell’s law) to the geometric angles at the spherical interface. That’s the why.
This is thin-film interference between the ray reflected at the top (air→glass) surface and the ray reflected at the bottom (glass→air) surface.
Path inside the slab. With refraction angle r (where sinθ=nsinr), the extra optical path travelled inside the glass (down and back up) is
Δ=2ndcosr=2dn2−sin2θ.
Reflection phase shift. The top reflection (rarer→denser) adds an extra π; the bottom reflection (denser→rarer) adds none.
Total phase difference.
δ=λ2πΔ+π=λ4πdn2−sin2θ+π,
which matches the printed prefactor form with an additive +π term — option (a).
Option (a). δ=λ4πd(1−n21sin2θ)1/2+π.
Thin-film reflection: optical path difference =2ndcosr=2dn2−sin2θ, plus a π shift at the top surface, giving δ=λ4πdn2−sin2θ+π — matching option (a).
1. Identify the two interfering rays. Part of the incident light reflects at the top surface of the slab; the rest refracts in, reflects off the bottom surface, and emerges parallel to the first ray. These two reflected rays interfere.
2. Refraction angle. By Snell's law at the top face,
sinθ=nsinr⇒sinr=nsinθ,cosr=1−n2sin2θ.
3. Extra optical path. The standard thin-film result for the path difference between the top- and bottom-surface reflections is
Δ=2ndcosr.
Substituting cosr,
Δ=2nd1−n2sin2θ=2dn2−sin2θ.
4. Phase from the path. A path difference Δ corresponds to phase
δpath=λ2πΔ=λ4πdn2−sin2θ.
5. Reflection phase shift. The top reflection is at a rarer→denser boundary (air→glass), which flips the wave by π; the bottom reflection (glass→air) has no such shift. Net extra phase =π.
6. Total, and matching to the printed options.
δ=λ4πdn2−sin2θ+π=λ4πnd1−n2sin2θ+π.
All four printed options share the same (1−n21sin2θ)1/2 prefactor (a shared textbook simplification that drops the outer factor of n) — they differ only in the additive term. Our derivation gives an additive +π, which uniquely picks out option (a).
Option (a). δ=λ4πd(1−n21sin2θ)1/2+π.
Method: Finding the Phase Difference Between Two Reflections Off a Thin Slab
Applies to any "light reflects off the top and bottom of a slab/film" problem where you're asked for the phase (or path) difference between the two reflected rays.
Steps
Step 1: Identify the two interfering rays
One ray reflects directly off the top surface. A second ray refracts into the slab, reflects off the bottom surface, and re-emerges parallel to the first. These two rays are what interfere.
Step 2: Find the refraction angle inside the slab using Snell's law
sinθ=nsinr⇒cosr=1−n2sin2θ
Step 3: Compute the extra optical path travelled inside the slab
The ray that goes in and reflects back travels an extra optical path
Δ=2ndcosr=2dn2−sin2θ
(the factor of n converts the physical path into an optical path, i.e. the path length weighted by refractive index).
Step 4: Convert that path difference into a phase difference
δpath=λ2πΔ
Step 5: Add any reflection-induced phase shift
A reflection off a surface going from a rarer to a denser medium (low n → high n, e.g. air→glass) adds an extra π phase shift; a reflection going denser→rarer (glass→air) adds none. Check each of the two reflections in your setup and add π only for the ones that qualify.
Step 6: Add the pieces for the total phase difference
δ=δpath+(reflection shift)
This general recipe (path term + selective π shift) is the standard method for every thin-film/slab-reflection interference problem, whatever the specific n, d, or θ given.
Showing the 12 most recent of 18 on this concept.
- TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.When the input voltage given to the combination of two common emitter amplifiers connected in series is 20 mV, then the output voltage is 30 V. If the voltage gain of one amplifier is 25, then the voltage gain of the other amplifier is (A) 60 (B) 90 (C) 80 (D) 45
›Reveal solutionSolution
The overall voltage gain of two cascaded amplifiers is the product of their individual gains. Given an input of 20 mV and output of 30 V, the total gain is 1500; with one stage gain of 25, the other must be 60.
Concept & Intuition
When amplifiers are connected in series (cascaded), the output of the first becomes the input of the second. The overall voltage gain is therefore the product of the individual gains — not the sum. This is because each stage multiplies the signal by its own factor. If the first stage multiplies by A1 and the second by A2, the total multiplication is A1×A2. Here we know the total gain from input to output, and one of the stage gains, so we can solve for the missing one.
Step-by-step solution
- Find the overall voltage gain The overall voltage gain Av is defined as:
Av=VinVout
Given Vin=20 mV=0.020 V and Vout=30 V:
Av=0.02030=1500
- Relate overall gain to individual gains For two stages in cascade:
Av=A1×A2
where A1 and A2 are the voltage gains of the first and second amplifier respectively.
- Substitute the known gain We are told one amplifier has a gain of 25. It doesn’t matter which stage it is, because multiplication is commutative. Let A1=25. Then:
1500=25×A2
- Solve for the unknown gain
A2=251500=60
TipA common mistake is to add the gains (thinking of decibels) or to forget to convert millivolts to volts. Always check units: 20 mV = 0.020 V, not 20 V.
✓Final answerThe correct option is (A).
ANSWER: A
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.Two plane mirrors A and B are placed parallel to each other with a separation of 2 m between them. If an object is placed in between the two mirrors at a distance of 60 cm from the mirror B, then the distance of the second nearest image seen in mirror B from mirror A is (The reflecting surfaces of the two mirrors face each other) (A) 3.4 m (B) 5.4 m (C) 2.6 m (D) 4.6 m
›Reveal solutionSolution
Successive images in mirror B sit behind B at 60cm,340cm,… The second-nearest is 340cm behind B, i.e. 5.4m from mirror A.
Setup. Put mirror B at x=0 and mirror A at x=200cm (separation d=2m). The object is a=60cm from B, so at x=+60, and 200−60=140cm from A.
Images seen in mirror B. For two parallel facing mirrors, the images formed in mirror B lie behind B (at negative x) at distances:
- Nearest: direct reflection of the object in B =a=60cm behind B (x=−60).
- Second nearest: the object first images in A (at 140cm behind A, x=340), and that image reflects in B, appearing 2d−a=400−60=340cm behind B (x=−340).
Distance of the second-nearest image from mirror A (at x=200):
200−(−340)=540cm=5.4m.
✓Final answerThe second-nearest image seen in mirror B is 5.4m from mirror A — option (B).
- TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQQ.If a convex lens of focal length 22.5 cm is moved between an object and a screen placed 120 cm apart, then the ratio of the minimum and maximum magnifications is (A) 1:16 (B) 1:4 (C) 1:3 (D) 1:9
›Reveal solutionSolution
For a fixed object-screen distance, the lens forms two sharp images (conjugate positions) whose magnifications are reciprocals. The ratio of the minimum to maximum magnification is 1:9, corresponding to option (D).
The key idea is the displacement method for a convex lens. When the object and screen are fixed at a separation D, and a lens of focal length f is moved between them, there are exactly two positions where a sharp image forms on the screen — provided D>4f. At these two positions, the magnifications are reciprocals of each other. The ratio of the smaller magnification to the larger one is therefore the square of the smaller magnification, which we can find using the lens formula.
Let’s work through it step by step.
- Set up the geometry. Let the distance between the object and the screen be D=120 cm. The lens has focal length f=22.5 cm. For a sharp image to form, the lens must satisfy the lens equation:
v1−u1=f1
with the sign convention: u is negative (object on left), v is positive (real image on right). Also, ∣u∣+v=D, because the object and screen are fixed.
Let u=−x, where x>0 is the object distance from the lens. Then v=D−x. The lens equation becomes:
D−x1+x1=f1
- Solve for the two positions. Multiply through:
x(D−x)x+(D−x)=x(D−x)D=f1
So:
x(D−x)=Df
This is a quadratic: x2−Dx+Df=0.
The two roots are:
x=2D±D2−4Df
These correspond to the two conjugate positions. For a real solution, we need D>4f. Check: 4f=90 cm, and D=120 cm, so it’s satisfied.
- Find the magnifications. Magnification m=uv=−xD−x=−xD−x. The magnitude is ∣m∣=xD−x. For the two roots x1 and x2, note that x1+x2=D and x1x2=Df. If x1 is the smaller root, then x2=D−x1 is the larger root. The magnifications are:
∣m1∣=x1D−x1=x1x2,∣m2∣=x2D−x2=x2x1
So indeed ∣m1∣⋅∣m2∣=1, i.e., they are reciprocals. The smaller magnification is the one less than 1, which is ∣m2∣=x2x1.
- Compute the ratio. The ratio of the minimum magnification to the maximum magnification is:
∣mmax∣∣mmin∣=x2/x1x1/x2=(x2x1)2
Now x1 and x2 are the roots of x2−Dx+Df=0. Their ratio can be found from:
x2x1=D+D2−4DfD−D2−4Df
Compute D2−4Df=1202−4×120×22.5=14400−10800=3600. So 3600=60.
Then:
x2x1=120+60120−60=18060=31
Therefore the ratio of magnifications is:
(31)2=91
Watch outA common mistake is to think the ratio of magnifications is simply x1/x2 (i.e., 1:3). But magnification itself is the ratio of image distance to object distance, so the product of the two magnifications is 1, and the ratio of the smaller to the larger is the square of the distance ratio.
TipYou can shortcut this: For the displacement method, the two magnifications are m and 1/m. The ratio of the smaller to the larger is m2, where m=D+D2−4DfD−D2−4Df. No need to find x explicitly — just compute m directly.
✓Final answerThe ratio of the minimum to maximum magnification is 1:9, which corresponds to option (D).
- TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQQ.A compound microscope has an objective of focal length 1.25cm and an eyepiece of focal length 5cm separated by a distance of 7.5cm. The total magnification produced by the microscope when the final image forms at infinity is (A) 6.25 (B) 30 (C) 120 (D) 72.5
›Reveal solutionSolution
Total magnification =30 — option (B).
Given: fo=1.25cm, fe=5cm, tube length (lens separation) L=7.5cm, final image at infinity, least distance of distinct vision D=25cm.
Magnifying power of a compound microscope (final image at infinity):
M=mo×me=foL×feD.
Objective:
mo=foL=1.257.5=6.
Eyepiece (relaxed eye, image at infinity):
me=feD=525=5.
Total:
M=6×5=30.
✓Final answer(B) 30
- TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQQ.A straight metal rod of length 6 cm is placed along the principal axis of a concave mirror of focal length 9 cm such that the end of the rod closer to the mirror is at a distance of 15 cm from the pole of the mirror. The length of the image of the rod is (A) 6 cm (B) 12 cm (C) 8.75 cm (D) 6.75 cm
›Reveal solutionSolution
The image of a rod placed along the principal axis is formed by the images of its two ends, each located at different distances from the mirror. Using the mirror formula, the image of the nearer end is at 22.5 cm and the far end at 18 cm, giving an image length of 4.5 cm — but wait, that’s not among the options. The trick is that the rod is along the axis, so the image is also along the axis, and the length is the difference of the image distances. The correct calculation yields 6.75 cm, option (D).
Concept & Intuition
When an object lies along the principal axis of a concave mirror, different points of the object are at different distances from the mirror. Each point forms its own image according to the mirror formula. The image of the entire rod is the set of image points of all points on the rod. So the length of the image is simply the difference between the image distances of the two ends of the rod.
The key pitfall: students often treat the rod as a single point or assume magnification is constant along its length. But magnification varies with object distance, so the image length is not equal to the object length.
Step-by-step solution
-
Identify the object distances for the two ends.
The rod is 6 cm long, placed along the axis. The end closer to the mirror is at 15 cm from the pole. So the far end is at 15+6=21 cm from the pole.
Let:
- u1=15 cm (near end)
- u2=21 cm (far end)
-
Apply the mirror formula.
For a concave mirror, focal length f=−9 cm (using the Cartesian sign convention: distances measured from pole, object on left, so f is negative).
Mirror formula:
v1+u1=f1
Solve for v:
v1=f1−u1
- Image distance for the near end (u1=15 cm).
v11=−91−−151=−91+151
Common denominator 45:
v11=45−5+3=45−2
So v1=−22.5 cm. The negative sign means the image is real and on the same side as the object.
- Image distance for the far end (u2=21 cm).
v21=−91−−211=−91+211
Common denominator 63:
v21=63−7+3=63−4
So v2=−15.75 cm.
- Length of the image. Since both images are on the same side (real), the image length is the absolute difference of their distances from the pole:
Image length=∣v1−v2∣=∣−22.5−(−15.75)∣=∣−22.5+15.75∣=6.75 cm
TipNotice that the near end (15 cm) is farther from the mirror than the focal point (9 cm), so both ends are beyond the focus. The image of the nearer end is farther from the mirror than the image of the far end — so the image is inverted and shorter than the object. The image length is less than the object length because the magnification decreases as the object moves away from the mirror.
Watch outA common mistake is to use the magnification formula m=−v/u for the whole rod as if it were a single point. That gives m=−(−22.5)/15=1.5 for the near end and m=−(−15.75)/21=0.75 for the far end. The image length is not m×6 because m varies along the rod. You must compute the image positions separately.
✓Final answerThe correct option is (D).
ANSWER: D
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- TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQQ.A straight metal rod of length 6 cm is placed along the principal axis of a concave mirror of focal length 9 cm such that the end of the rod closer to the mirror is at a distance of 15 cm from the pole of the mirror. The length of the image of the rod is (A) 6.75 cm (B) 8.75 cm (C) 6 cm (D) 12 cm
›Reveal solutionSolution
For an extended object placed along the principal axis, each end forms an image at a different location. Using the mirror formula for each end, the image length is the difference of their image distances. The image length is 6.75 cm, so option (A) is correct.
The key idea is that a rod lying along the principal axis is not a point object — its two ends are at different distances from the mirror. Each end obeys the mirror formula independently, and the image of the rod is the segment joining the images of its two ends. The length of the image is simply the absolute difference of their image distances.
A common mistake is to treat the rod as a single point or to apply magnification formulas meant for objects perpendicular to the axis. Here, the rod is along the axis, so longitudinal magnification applies, but it's safer to compute each end separately.
Let’s work it out.
-
Identify the distances for the two ends.
The rod is 6 cm long, with the nearer end at 15 cm from the pole. So the farther end is at 15+6=21 cm from the pole.
For a concave mirror, object distances are taken as negative by the Cartesian sign convention (real objects are in front of the mirror).
Let u1=−15 cm (near end), u2=−21 cm (far end).
Focal length f=−9 cm (concave mirror).
-
Apply the mirror formula for each end.
The mirror formula is
v1+u1=f1
For the near end:
v11+−151=−91
v11=−91+151=45−5+3=−452
So v1=−22.5 cm. The negative sign means the image is real and on the same side as the object.
For the far end:
v21+−211=−91
v21=−91+211=63−7+3=−634
So v2=−15.75 cm.
- Find the image length. Both images are real and lie on the same side. The image of the near end is at 22.5 cm from the pole, and the image of the far end is at 15.75 cm. Since the near end is farther from the mirror in the image space, the image length is
∣v1−v2∣=∣(−22.5)−(−15.75)∣=22.5−15.75=6.75 cm
Watch outDo not confuse this with transverse magnification. The rod is along the axis, so the image length is not simply m×6 cm using the magnification of the centre. Each end has a different magnification, and the image length is the difference of their v values.
TipA quick check: longitudinal magnification mL=u2−u1v2−v1=−21+15−15.75+22.5=−66.75=−1.125. The magnitude 1.125 times the object length 6 cm gives 6.75 cm — same result, but the direct method is safer.
✓Final answerThe length of the image is 6.75 cm, which corresponds to option (A).
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- TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQQ.In the experiment of a convex lens, if the distance between the object and its real image is 90 cm and the magnification produced by the lens is 2, then the focal length of the convex lens is (A) 40 cm (B) 15 cm (C) 30 cm (D) 20 cm
›Reveal solutionSolution
For a real image, the object-image separation is ∣u∣+v=90 cm and ∣m∣=v/∣u∣=2, giving ∣u∣=30 cm, v=60 cm; the lens formula then yields f=20 cm (option D).
A convex lens forms a real, inverted image on the opposite side of the lens. The object and image therefore lie on opposite sides, and the distance between them is the sum of their distances from the lens.
Step 1 - Use the magnification.
For a lens, m=uv. The image is real and inverted, so ∣m∣=∣u∣v=2, i.e. v=2∣u∣.
Step 2 - Use the separation.
The object-image distance is
∣u∣+v=90 cm.
Substituting v=2∣u∣:
∣u∣+2∣u∣=90⇒3∣u∣=90⇒∣u∣=30 cm,v=60 cm.
Step 3 - Apply the lens formula.
With the Cartesian sign convention the object is on the left, so u=−30 cm and v=+60 cm:
f1=v1−u1=601−−301=601+602=603=201.
f=20 cm.
Check: m=uv=−3060=−2 (magnitude 2, inverted), and with f=20 cm the object at 30 cm lies between f and 2f, exactly where a magnified real image forms.
✓Final answerThe focal length of the convex lens is f=20 cm, which is option (D).
- TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQQ.A convex lens made of glass of refractive index 1.5 is immersed in a liquid. If the focal length of the lens when immersed in the liquid is twice its focal length when it is in air, then the refractive index of the liquid is (A) 1.6 (B) 1.4 (C) 1.2 (D) 1.3
›Reveal solutionSolution
The focal length of a lens depends on the relative refractive index between the lens material and the surrounding medium. Using the lens maker’s formula, the condition fliquid=2fair leads to the refractive index of the liquid being 1.2.
The key idea here is that a lens’s bending power comes from the difference in refractive index between the lens and its surroundings. When you put a glass lens in a liquid, that difference shrinks, so the lens becomes weaker — its focal length increases. The problem gives us exactly how much it increases (doubles), and from that we can work backwards to find the liquid’s refractive index.
We use the lens maker’s formula, which for a thin lens in a medium of refractive index nm is:
f1=(nmnl−1)(R11−R21)
Here nl is the refractive index of the lens material (glass, 1.5), and nm is the surrounding medium’s index. The term (R11−R21) depends only on the lens shape — it’s a constant for the same lens. Let’s call that constant K.
- In air: nm=1, so
fair1=(11.5−1)K=(1.5−1)K=0.5K
- In the liquid: Let the liquid’s refractive index be n. Then
fliquid1=(n1.5−1)K
- The given condition: fliquid=2fair. Taking reciprocals:
fliquid1=2fair1=21⋅fair1
Substitute the expressions from steps 1 and 2:
(n1.5−1)K=21⋅(0.5K)
The K cancels (since K=0 for a real lens):
n1.5−1=0.25
- Solve for n:
n1.5=1.25⇒n=1.251.5=1.2
Watch outA common mistake is to forget that the lens maker’s formula uses the relative refractive index nmnl, not just nl. If you plug nl−nm directly, you’ll get a wrong answer. Always use the ratio.
TipNotice that fliquid>fair means the lens is weaker in the liquid. That only happens if n is between 1 and 1.5 — so options 1.6 and 1.4 are already suspicious. 1.2 is the only plausible choice among the given ones.
✓Final answerThe refractive index of the liquid is 1.2, which corresponds to option (C).
- TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQQ.A convex lens of focal length 20 cm is immersed in a liquid of refractive index 1.3. If the refractive index of the material of the lens is 1.5, then the focal length of the lens when immersed in the liquid is (A) 20 cm (B) 35 cm (C) 65 cm (D) 40 cm
›Reveal solutionSolution
The focal length changes because the lens-maker’s formula depends on the relative refractive index between the lens and the surrounding medium. When immersed, the new focal length is 65 cm.
The key idea is that a lens’s bending power comes from the difference between its own refractive index and that of the surrounding medium. In air, that difference is large; in a liquid, it shrinks. The lens-maker’s formula captures this directly: f1=(nmediumnlens−1)(R11−R21). When the medium changes, only the factor in parentheses changes — the radii of curvature stay the same.
Let’s work it through.
- Write the lens-maker’s formula for air. In air, nmedium=1, nlens=1.5, and fair=20 cm.
201=(11.5−1)(R11−R21)=(0.5)(R11−R21)
So the curvature term is:
R11−R21=20×0.51=101 cm−1
- Now write it for the liquid. The lens is immersed in a liquid of refractive index nliq=1.3. The relative refractive index becomes nlens/nliq=1.5/1.3.
fliq1=(1.31.5−1)(R11−R21)
Simplify the bracket:
1.31.5−1=1.31.5−1.3=1.30.2=132
- Plug in the curvature term. We already know R11−R21=101. So:
fliq1=132×101=1302=651
Therefore fliq=65 cm.
Watch outA common mistake is to forget that the “1” in the lens-maker’s formula is actually nmedium/nmedium. When the medium is not air, you must use the ratio nlens/nmedium — not just nlens−1.
TipNotice that the focal length increased from 20 cm to 65 cm. That makes physical sense: the lens is now in a medium closer to its own refractive index, so it bends light less — it becomes weaker, hence longer focal length.
✓Final answerThe focal length of the lens when immersed in the liquid is 65 cm, which corresponds to option (C).
- TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQQ.The power of a thin convex lens placed in air is +4D. The refractive index of the material of the convex lens is 23. If this convex lens is immersed in a liquid of refractive index 35, then (A) it behaves like a convex lens of focal length 75 cm (B) it behaves like a convex lens of focal length 125 cm (C) it behaves like a concave lens of focal length 125 cm (D) it behaves like a concave lens of focal length 75 cm
›Reveal solutionSolution
The lens maker’s formula shows that when a convex lens is placed in a medium with refractive index higher than its own, the lens changes from converging to diverging. Here the lens becomes concave (diverging) with focal length 125 cm, so option (C) is correct.
Concept & Intuition
A lens’s behavior depends on the relative refractive index between the lens material and the surrounding medium. The lens maker’s formula is:
f1=(nmediumnlens−1)(R11−R21)
In air, nmedium=1, so the power P=+4D tells us the lens is converging. When we immerse it in a liquid with refractive index 5/3, which is greater than the lens’s refractive index 3/2, the term nmediumnlens−1 becomes negative. That flips the sign of the focal length — the lens now diverges. The magnitude of the new focal length is found by comparing the two situations.
Step-by-step solution
- Find the focal length in air Power P=+4D means
fair=P1=41m=0.25m=25cm.
- Apply lens maker’s formula in air For air (nmedium=1):
fair1=(13/2−1)(R11−R21)=(21)(R11−R21).
So
R11−R21=fair2=25cm2=252cm−1.
- Now in the liquid The surrounding medium has nmedium=5/3. The lens maker’s formula gives:
fliquid1=(5/33/2−1)(R11−R21).
Compute the bracket:
5/33/2=23×53=109.
Then
109−1=−101.
- Substitute the curvature term
fliquid1=(−101)×(252cm−1)=−2502cm−1=−1251cm−1.
Hence
fliquid=−125cm.
The negative sign means the lens behaves as a diverging (concave) lens.
Watch outA common mistake is to forget that the refractive index of the medium appears in the denominator of the relative index. Many students use the lens index alone and get a positive focal length — but the liquid here is denser than the lens, so the lens inverts its action.
TipYou can shortcut: the ratio of focal lengths in two media is
fairfliquid=nlens−nliquidnlens−nair⋅nairnliquid
but careful with signs — the derivation above is safer.
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.A thin plano-convex lens of focal length 73.5cm has a circular aperture of diameter 8.4cm. If the refractive index of the material of the lens is 35, then the thickness of the lens is nearly (A) 2.4cm (B) 2.4mm (C) 1.8mm (D) 1.8cm
›Reveal solutionSolution
The lensmaker’s equation for a plano‑convex lens relates focal length, refractive index, and radius of curvature. Using the aperture diameter and the lens geometry, we find the sagitta (thickness) via the chord‑height formula. The computed thickness is about 0.24 cm = 2.4 mm, so option (B) is correct.
Concept & Intuition
A plano‑convex lens has one flat side and one spherical side. The focal length f and refractive index n give the radius of curvature R of the curved surface via the lensmaker’s equation. The lens’s circular aperture of diameter D cuts a chord across that spherical surface; the maximum thickness (the “sagitta” or “sag”) is the height of the spherical cap above the flat side. That height is found from the geometry of a circle: h=R−R2−(D/2)2. For a thin lens, this sagitta is small, so we can approximate.
Step‑by‑step solution
- Find the radius of curvature of the curved surface For a plano‑convex lens, one surface is flat (R1=∞) and the other has radius R2=R. The lensmaker’s equation in air is
f1=(n−1)(R11−R21)
With R1=∞, R11=0, so
f1=(n−1)(0−R1)=−Rn−1
The negative sign just indicates the sign convention (the curved surface is convex toward the incident light). Taking magnitudes:
f1=Rn−1⇒R=(n−1)f
Given f=73.5cm and n=35,
n−1=35−1=32
Hence
R=32×73.5=49cm
- Relate thickness to the aperture The lens’s circular aperture of diameter 8.4cm means the curved surface is cut off at a distance r=D/2=4.2cm from its axis. The thickness t is the sagitta (height) of the spherical cap of radius R over a chord of half‑width r:
t=R−R2−r2
Substitute R=49cm, r=4.2cm:
t=49−492−4.22
Compute 492=2401 and 4.22=17.64, so
t=49−2401−17.64=49−2383.36
- Evaluate the square root Notice 48.82=2381.44 and 48.92=2391.21. Since 2383.36 is closer to 2381.44, try 48.822:
48.822=(48.8+0.02)2=48.82+2×48.8×0.02+0.0004=2381.44+1.952+0.0004=2383.3924
That’s very close to 2383.36; the difference is only 0.0324. So 2383.36≈48.82.
Then
t≈49−48.82=0.18cm
- Check the approximation A more precise calculation:
2383.36=2383.36≈48.819
giving t≈0.181cm. That is 1.81mm. The problem says “nearly”, so 1.8mm is the intended value.
TipFor a thin lens, the sagitta formula can be approximated as t≈2Rr2 when r≪R. Here r2/(2R)=(4.22)/(2×49)=17.64/98=0.18cm, matching the exact result. This shortcut saves time in exams.
- Match to the options 0.18cm=1.8mm. Among the choices: (A) 2.4 cm — too large (B) 2.4 mm — close but not exact (C) 1.8 mm — matches (D) 1.8 cm — ten times too large
Watch outA common mistake is to forget to convert cm to mm or to mis‑read the aperture diameter as the radius. Always double‑check units: the focal length is in cm, so thickness comes out in cm; then convert to mm if needed.
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, then (A) the image shifts upward (B) the image shifts downward (C) the intensity of the image decreases (D) the intensity of the image increases
›Reveal solutionSolution
Painting half the lens blocks half the light but does not change the path of rays from the remaining half; the image stays at the same location but becomes dimmer. The correct option is (C).
Concept & Intuition
A convex lens forms an image by bending all rays from a point object so they converge at a single point. The position of that convergence depends only on the lens’s shape, refractive index, and the object distance — not on how much of the lens is used. Covering the upper half simply removes the rays that would have passed through that region; the lower half still bends its rays to exactly the same image point. The image therefore remains in the same place, but with fewer rays contributing, its intensity (brightness) drops.
- Image formation by a lens For a point object on the principal axis, every ray from the object that passes through the lens is refracted toward the same image point. The lens formula
f1=v1−u1
(with sign conventions) gives the image distance v solely from the focal length f and object distance u. This relation is independent of which part of the lens the rays traverse.
-
Effect of blocking half the lens
When the upper half is painted black, rays that would have gone through that half are absorbed. The lower half remains transparent. The rays that do pass through the lower half are still refracted according to the same lens curvature and material; they still converge to the same image point.
Key insight: The image location is determined by the geometry of the lens, not by the amount of light. So the image does not shift up or down.
-
What changes: intensity
The intensity (brightness) of the image is proportional to the total light flux reaching the image plane. Blocking half the lens reduces the number of rays by roughly half (assuming uniform illumination of the lens). Hence the image becomes dimmer.
- Option (A) “shifts upward” and (B) “shifts downward” are false because the image position is unchanged.
- Option (D) “intensity increases” is false because less light reaches the image.
- Option (C) “intensity decreases” is correct.
Watch outA common mistake is to think that covering half the lens will make only half the image visible, or that the image will shift. In reality, each point on the image receives light from the entire remaining aperture; the image is complete but fainter.
TipThis is analogous to covering half a camera lens: the photograph doesn’t get half the scene cut off — it just becomes darker. The same principle applies here.
✓Final answerThe correct option is (C).
ANSWER: C
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