(a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the ex

Question

Q. (a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope.

(b) An astronomical telescope has an objective lens of focal length 20 m and eyepiece of focal length 1 cm.

(i) Find the angular magnification of the telescope.

(ii) If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is

3{\cdot}5 \times 10^{6}

m and radius of lunar orbit is

3{\cdot}8 \times 10^{8}

(OR)

(a) An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation

\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

(b) An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 cm. If the refractive index of the material of the lens is

1{\cdot}5

, find the position and nature of the image formed.

CBSECBSE Class XII Board 2020Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Angular Magnification

What is Angular Magnification?

When you look at a tiny object — say a grain of salt — you hold it close to your eye to see it bigger. But there is a limit: bring it too close and it blurs. The closest distance at which your eye can focus comfortably is called the near point, conventionally taken as 25 cm for a normal eye. At that distance, the object subtends a certain angle at your eye. That angle determines how large it appears — not its physical size, but the fraction of your field of view it occupies.

Now imagine using a magnifying glass. The same grain of salt now looks much larger. Why? Because the lens lets you bring the object even closer than 25 cm while still seeing a clear, magnified image. That image is formed at a comfortable viewing distance, but the angle it subtends at your eye is far bigger than the angle the object would subtend at 25 cm without the lens.

Angular magnification is simply the ratio of these two angles:

Angular magnification $M = \frac{\theta_{\text{image}}}{\theta_{\text{object}}}$

where $\theta_{\text{image}}$ is the angle subtended by the image when viewed through the instrument, and $\theta_{\text{object}}$ is the angle subtended by the object when viewed with the naked eye at the near point (25 cm).

Why "Angular" and Not "Linear"?

A common confusion: a microscope or telescope does not give you a physically larger object — it gives you a larger apparent size. The image on your retina is bigger because the rays entering your eye are steeper. That steepness is measured by the angle. So magnification here is about angles, not actual lengths.

Important

Angular magnification is dimensionless. It tells you how many times wider the image appears compared to the object seen directly at the near point.

A Concrete Example

Take a simple magnifier (a convex lens) of focal length $f = 5$ cm. You place the object just inside the focal point so that a virtual, erect image forms at infinity (or at the near point). For the "image at infinity" case, the angle subtended by the image is $\theta_{\text{image}} \approx h / f$ , where $h$ is the object height. The angle subtended by the object at the near point (25 cm) is $\theta_{\text{object}} \approx h / 25$ .

Thus:

M = \frac{h/f}{h/25} = \frac{25}{f}

For $f = 5$ cm, $M = 5$ . The image appears 5 times larger than the object seen at 25 cm.

Tip

For a magnifier, the formula $M = 1 + \frac{25}{f}$ applies when the image is formed at the near point (25 cm) — giving slightly higher magnification than the infinity-focus case.

The Big Picture

Angular magnification is the language of all optical instruments:

Simple magnifier: $M \approx 25/f$ (image at infinity) …

Part (b)Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

Imagine you're standing in front of a concave mirror — the kind that makes your face look bigger when you're close, but flips everything upside down when you step far back. That change isn't magic; it's geometry. The spherical mirror equation is the single relationship that predicts exactly where an image will form, and whether it's real or virtual, for any spherical mirror.

The Core Idea

Every point on an object sends out light rays in all directions. A mirror redirects those rays. The mirror equation tells you: given the mirror's curvature and the object's distance, where will those rays meet again (or appear to meet)?

There are only three quantities you need:

$u$ — object distance (from the mirror's pole)
$v$ — image distance (from the mirror's pole)
$f$ — focal length (a property of the mirror's curvature)

The equation is:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

The power is in the sign convention, because every distance can point in one of two directions.

The Sign Convention (New Cartesian Sign Convention)

This is where most students slip. The equation works for all spherical mirrors — concave and convex — only if you follow the convention used throughout NCERT and CBSE:

All distances are measured from the mirror's pole.
The incident light is taken to travel left to right, so distances measured in that same direction (to the right) are positive, and distances measured against it (to the left) are negative.
Heights above the principal axis are positive; heights below are negative.

Because a real object is always placed in front of the mirror (to the left, where the incident light originates), its distance $u$ is always negative.

Important

Under this convention, the focal length of a concave mirror is negative (its focus $F$ sits in front of the mirror, on the same side as the object), and the focal length of a convex mirror is positive (its focus lies behind the mirror). This is one of the most frequently tested facts in CBSE board exams.

Watch out

A very common mistake is writing $f$ as positive for a concave mirror because "it converges light." Convergence tells you the type of mirror, not the sign — the sign comes purely from where the focus physically sits relative to the pole, under the convention above.

Where Does the Formula Come From?

For a concave mirror, parallel rays from a distant object converge at the focus, a point at (signed) distance $f$ from the mirror. The derivation uses similar triangles from a ray diagram.

›Proof

Consider an object of height $h_o$ in front of a concave mirror. Draw the ray parallel to the axis: it reflects through the focus $F$ . Draw the ray through the centre of curvature $C$ : it strikes the mirror normally and reflects straight back on itself. These two reflected rays cross to form the image, of height $h_i$ .

From similar triangles formed by the ray through $C$ :

$\frac{h_o}{h_i} = \frac{u - R}{R - v}$

where $R = 2f$ is the radius of curvature (with the same sign convention as $f$ ).

From similar triangles formed by the ray through $F$ :

$\frac{h_o}{h_i} = \frac{u - f}{f}$

Equating the two ratios and simplifying (using $R = 2f$ ) gives:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

What the Equation Tells You

Rearranging for $v$ :

v = \frac{uf}{u - f}

Because $u$ is negative for a real object, and $v$ takes the sign the geometry dictates:

$v$ negative → the image forms in front of the mirror → real image (can be projected on a screen).
$v$ positive → the image forms behind the mirror → virtual image.

For a concave mirror ( $f$ negative), using the magnitude of the object distance $|u|$ measured from the pole:

$|u| > 2|f|$ → real, inverted, diminished image between $f$ and $2f$
$|u| = 2|f|$ → real, inverted, same-size image at $2f$
$|f| < |u| < 2|f|$ → real, inverted, magnified image beyond $2f$
$|u| = |f|$ → image at infinity …

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

The spherical mirror equation — also called the mirror formula — relates the object distance ( $u$ ), image distance ( $v$ ), and focal length ( $f$ ) of a spherical mirror. Let's build the reasoning step by step.

1. The Key Formula

For a spherical mirror (concave or convex):

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Where:

$f$ = focal length (positive for concave, negative for convex)
$u$ = object distance from pole (always negative by sign convention)
$v$ = image distance from pole (sign depends on image location)

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Consider a concave mirror with:

Pole $P$
Centre of curvature $C$ (radius $R$ )
Focus $F$ (midpoint of $PC$ , so $f = R/2$ )

Take an object placed beyond $C$ . Draw two rays from the object's tip:

A ray parallel to the principal axis → reflects through $F$
A ray through $C$ → reflects back along itself

These rays meet at the image point.

Step 2: Use similar triangles

Let the object height be $h_o$ and image height be $h_i$ .

From the geometry of the ray through $C$ :

Triangle formed by object, $C$ , and axis is similar to triangle formed by image, $C$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - R}{R - v}

(Here $u$ and $v$ are distances from $P$ , with sign conventions applied later.)

Step 3: Use the parallel ray

From the ray parallel to the axis:

Triangle formed by object, $F$ , and axis is similar to triangle formed by image, $F$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - f}{f}

Step 4: Equate the two ratios

Since both ratios equal $h_o/h_i$ :

\frac{u - R}{R - v} = \frac{u - f}{f}

Step 5: Substitute $R = 2f$

For a spherical mirror, the focal length is half the radius of curvature:

R = 2f

Substitute:

\frac{u - 2f}{2f - v} = \frac{u - f}{f}

Step 6: Cross-multiply and simplify

Cross-multiply:

f(u - 2f) = (u - f)(2f - v)

Expand:

fu - 2f^2 = 2fu - uv - 2f^2 + fv

Cancel $-2f^2$ on both sides:

fu = 2fu - uv + fv

Bring all terms to one side:

0 = fu - uv + fv

Rearrange:

uv = fu + fv

Step 7: Divide by $uvf$

Divide both sides by $uvf$ :

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

This is the mirror formula.

3. Why the Sign Convention Matters

The derivation above used distances as positive magnitudes. In actual problem-solving, we use the Cartesian sign convention:

Distances measured against incident light are negative
Distances measured along incident light are positive

For a concave mirror:

$u$ is negative (object in front)
$f$ is negative (focus in front)
$v$ is negative for real images (in front) …

Start your 14-day free trial to unlock the full solution →

Part (a): telescope R.P. $=\dfrac{D}{1.22\lambda}$ ; angular magnification $M=f_o/f_e=2000$ and the Moon's image at the objective is $\approx 0.184$ m across.

Part (b): the mirror equation $\dfrac1v+\dfrac1u=\dfrac1f$ is derived for a concave mirror forming a virtual image; the plano-convex lens ( $f=40$ cm) forms a virtual, erect, $4\times$ image $120$ cm in front of the lens.

Ray diagram of an astronomical refracting telescope with the final image formed at infinity (normal adjustment): parallel rays from a distant object converge at the objective's focal point Fo, which coincides with the eyepiece's focal point Fe, so the rays emerge from the eyepiece parallel and the final, inverted image is at infinity for a relaxed eye.

Part (a): Astronomical Telescope and Resolving Power

Ray diagram (normal adjustment, image at infinity).

Draw the principal axis with the objective $L_o$ on the left and eyepiece $L_e$ on the right, separated by $f_o+f_e$ .
Parallel rays from a distant object arrive making a small angle $\alpha$ with the axis; the objective forms a real, inverted, diminished intermediate image $A'B'$ at its focal point $F_o$ .
$L_e$ is positioned so its focal point $F_e$ coincides with $A'B'$ ; rays leave the eyepiece parallel, making a larger angle $\beta$ with the axis — the final image is inverted and at infinity.

Resolving power (Rayleigh criterion for a circular aperture):

\text{R.P.} = \frac{1}{\Delta\theta} = \frac{D}{1.22\,\lambda}

Numerical. Given $f_o=20$ m, $f_e=1$ cm $=0.01$ m, $D_M=3.5\times10^6$ m, $R_L=3.8\times10^8$ m.

Angular magnification: $M=\dfrac{f_o}{f_e}=\dfrac{20}{0.01}=2000$ .
Angle subtended by the Moon at the objective: $\alpha\approx\dfrac{D_M}{R_L}$ . The image sits at the objective's focal plane, so $\alpha\approx\dfrac{d_i}{f_o}$ . Equating, …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$144$
›Reveal solutionSolution
For a telescope in normal adjustment, the magnifying power is the ratio of objective to eyepiece focal lengths ( $M = f_o / f_e$ ), and the tube length is their sum ( $L = f_o + f_e$ ). Here, $M = 144/6 = 24$ and $L = 144 + 6 = 150$ cm, so the correct option is (A).

The question gives you a telescope with an objective of focal length $f_o = 144$ cm and an eyepiece of focal length $f_e = 6.0$ cm. You need the magnifying power and the length of the telescope tube.

The key idea is that a telescope is used to view distant objects. Light from a faraway object arrives as nearly parallel rays. The objective lens forms a real, inverted image at its focal plane. The eyepiece then acts as a magnifier to view that image.

For relaxed viewing (normal adjustment), the eyepiece is adjusted so that the final image is at infinity. This means the image formed by the objective must lie exactly at the focal point of the eyepiece. So the distance between the two lenses — the tube length — is simply the sum of their focal lengths.

The magnifying power (or angular magnification) is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended by the object when viewed directly. For a telescope in normal adjustment, this ratio simplifies beautifully to the ratio of the focal lengths.
$M = \frac{f_o}{f_e} \quad \text{and} \quad L = f_o + f_e$
Let's apply this directly.

Magnifying power:

$M = \frac{f_o}{f_e} = \frac{144}{6.0} = 24$

Tube length:

$L = f_o + f_e = 144 + 6.0 = 150 \text{ cm}$
So the magnifying power is 24 and the tube length is 150 cm. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. The location of the image is$
(a) formed at 6.67 cm behind the mirror.
(b) formed at 67 cm behind the mirror.
(c) formed at 70 cm same side of the mirror.
(d) formed at 5.57 cm same side of the mirror.
›Reveal solutionSolution

Using the mirror formula with the correct sign convention, the convex mirror forms a virtual image 6.67 cm behind the mirror.

Given: Needle height $h=4.5$ cm, object distance $u = -12$ cm (object in front, so negative by convention), convex mirror so focal length $f = +15$ cm (behind the mirror, positive).

Mirror formula:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$

Taking LCM (60): $\dfrac{1}{v} = \dfrac{4}{60} + \dfrac{5}{60} = \dfrac{9}{60} = \dfrac{3}{20}$

$v = \frac{20}{3} = 6.67\ \text{cm}$ …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/4/11 markMCQ
$-2.0$
›Reveal solutionSolution
A magnification of $-2.0$ means the image is inverted (negative sign) and magnified (magnitude > 1). Only a concave mirror can produce an inverted, magnified image, and such an image is always real. So the mirror is concave and the image is real — option (B).

Concept and Intuition

The magnification $m$ of a spherical mirror tells you two things at once: the sign tells you orientation, and the magnitude tells you size.

If $m$ is positive, the image is virtual and erect (upright).

If $m$ is negative, the image is real and inverted (upside down).

The magnitude $|m|$ tells you relative size:

$|m| > 1$ → image is magnified (larger than object)

$|m| < 1$ → image is diminished (smaller)

$|m| = 1$ → same size

Here $m = -2.0$ means the image is inverted (negative) and twice as large as the object ( $|m| = 2$ ).

Now, which mirror can produce an inverted, magnified image? A convex mirror always gives a virtual, erect, and diminished image — so it can never produce a negative magnification. A concave mirror, however, can produce both real (inverted) and virtual (erect) images depending on where the object is placed. The real image from a concave mirror is always inverted, and when the object is between the centre of curvature and the focus, that real image is also magnified.

So the only mirror that fits $m = -2.0$ is a concave mirror, and the image must be real.

Step-by-step reasoning

Interpret the sign of $m$

$m = -2.0$ is negative. For spherical mirrors, a negative magnification always means the image is inverted relative to the object. An inverted image formed by a single mirror is always real (it can be projected on a screen). So the image is real.

Interpret the magnitude of $m$

$|m| = 2.0 > 1$ , so the image is magnified — larger than the object.

Eliminate convex mirror

A convex mirror always produces a virtual, erect, and diminished image for any real object. That means $m$ is always positive and $|m| < 1$ . Since our $m$ is negative and $|m| > 1$ , a convex mirror is impossible. This eliminates options (A) and (D).

Check concave mirror possibilities

A concave mirror can produce:

A real, inverted, magnified image when the object is placed between $F$ and $C$ (focus and centre of curvature).

A virtual, erect, magnified image when the object is placed between $P$ and $F$ (pole and focus). In that case $m$ is positive.

Since our $m$ is negative, the image cannot be virtual. So the only possibility is the real, inverted, magnified case — which is exactly what a concave mirror gives for an object between $F$ and $C$ . …
Start your 14-day free trial to see the full solution →
CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. Assertion (A): The radius of curvature of a concave mirror is 20 cm. If an object is placed in front of the mirror at a distance of 10 cm from its pole, its image is formed at infinity. Reason (R): The image of an object placed at the focus of a spherical mirror is formed at infinity. Select the correct option.$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Both statements are true, and the reason correctly explains why the assertion is true.

For a concave mirror of radius of curvature $R = 20\,\text{cm}$ , the focal length is $f = R/2 = 10\,\text{cm}$ . In Assertion (A), the object is placed at a distance of 10 cm from the pole — exactly at the focus. Using the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ , when $u = f$ , we get $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{f} = 0$ , so $v \to \infty$ — the image is indeed formed at infinity. This is exactly the general principle stated in Reason (R): rays from an object placed at the f …

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$Q. Fill in the blank with appropriate word: Simple microscope is a converging lens of ______ focal length.$
›Reveal solutionSolution

A simple microscope must have a short focal length to give useful magnification.

A simple microscope is just a single convex (converging) lens used to view a small object placed within its focal length, forming a magnified, virtual, erect image. Its angular magnification (when the image is formed at the near point D = 25 cm) is given by m = 1 + D/f. This shows that the magnification increases as the focal length f decre …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of the lenses are$
(i) 10 cm, 20 cm
(ii) 15 cm, 5 cm
(iii) 18 cm, 2 cm
(iv) 11 cm, 9 cm
›Reveal solutionSolution

fo = 18 cm and fe = 2 cm.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Focal length of a concave mirror in air is 25 cm. Its focal length in water will be -$
(a) 50 cm
(b) 12.5 cm
(c) $\infty$
(d) 25 cm
›Reveal solutionSolution

A mirror's focal length depends only on its radius of curvature, not on the surrounding medium.

For a spherical mirror, $f = R/2$ , where $R$ is the radius of curvature -- a purely geometrical quantity. Since reflection (unlike refraction) does not depend on the refractive index of the surrounding medium, the focal length …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 markMCQ
$Q. The correct relationship between the radius of curvature (R) and focal length$
(f) of a spherical mirror is ______.
(a) R = 2f
(b) f = 2R
(c) R = f/2
(d) R = 1/f
›Reveal solutionSolution

For a spherical mirror, R = 2f because the focal point lies midway between the pole and the centre of curvature.

For a spherical mirror (concave or convex), a ray parallel to the principal axis, after reflection, passes through (or appears to diverge from) the focus F. Using the mirror geometry, for paraxial rays, the focal length f is related to the radius of curvature R by: …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. The radius of curvature of a concave mirror is 24 cm. The value of its focal length will be __________ cm.$
›Reveal solutionSolution

For a spherical mirror, f = R/2, a direct geometric consequence of paraxial ray reflection.

For any spherical mirror (concave or convex), the focal length is related to the radius of curvature by:

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. What should be increased to increase the angular magnification of a simple microscope?$
(a) The power of the lens
(b) The focal length of the lens
(c) Lens aperture
(d) Object size
›Reveal solutionSolution

Angular magnification $m=D/f$ (or $1+D/f$ ), so increasing the power ( $1/f$ ) directly increases $m$ .

For a simple microscope (magnifying glass), the angular magnification is $m = \dfrac{D}{f}$ (normal adjustment, image at infinity) or $m=1+\dfrac{D}{f}$ (image at the near point), where $D$ is the least distance of distinct vision and $f$ is the focal length. Since $m \propto \dfrac{1}{f}$ , and the power of a lens is $P=\dfrac{1}{f}$ , increasing the power of the lens (equivalently, decreasing $f$ ) directly increases the angular magnification. Increasing lens aperture …

Start your 14-day free trial to see the full solution →
CBSE 2023Set 55/1/11 markMCQ
$f$
(a) zero
(b) $f$
(c) $2f$
(d) $4f$
›Reveal solutionSolution
For a concave mirror, the object and its real image can be brought arbitrarily close together, but the minimum possible distance between them is zero — achieved when the object is at the centre of curvature and the image coincides with it.

The question asks for the minimum distance between an object and its real image formed by a concave mirror. This is a classic problem that tests your understanding of the mirror formula and the concept of real images.

The core idea

A real image is formed when rays actually converge after reflection. For a concave mirror, a real image is formed only when the object is placed beyond the focus (i.e., $u > f$ ). The image distance $v$ is then positive (real) and given by the mirror formula:
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
The distance between the object and its image is $|u - v|$ . We want to find the smallest possible value of this distance for real images.

Step-by-step reasoning

Set up the mirror formula For a concave mirror, $f$ is positive. Let $u$ be the object distance (positive, measured from the mirror). For a real image, $v$ is also positive. The mirror formula gives:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf}$
So:
$v = \frac{uf}{u - f}$

Write the distance between object and image Let $D = |u - v|$ . Since both $u$ and $v$ are positive and measured from the mirror, the object and image lie on the same side of the mirror. The distance between them is:

$D = |u - v| = \left| u - \frac{uf}{u - f} \right|$

Simplify the expression Factor $u$ :

$D = u \left| 1 - \frac{f}{u - f} \right| = u \left| \frac{u - f - f}{u - f} \right| = u \left| \frac{u - 2f}{u - f} \right|$
Since for real images $u > f$ , the denominator $u - f > 0$ . The sign of $u - 2f$ depends on $u$ . So:
$D = u \cdot \frac{|u - 2f|}{u - f}$

Analyse the behaviour

If $u > 2f$ , then $u - 2f > 0$ , so $D = u \cdot \frac{u - 2f}{u - f}$ .

If $f < u < 2f$ , then $u - 2f < 0$ , so $D = u \cdot \frac{2f - u}{u - f}$ .

In both cases, $D$ is positive. The question is: can $D$ become zero?

When does $D = 0$ ?

$D = 0$ when $|u - 2f| = 0$ , i.e., when $u = 2f$ . …
Start your 14-day free trial to see the full solution →
CBSE 2023Set F1 markMCQ
$Q. A spherical mirror is immersed in water. Its focal length will (A) decrease (B) increase (C) remain same (D) none of these$
›Reveal solutionSolution

A mirror's focal length depends only on its radius, so immersing it in water does not change $f$ .

Unlike a lens (whose focal length depends on the refractive index of the medium through the lens-maker's formula), a spherical mirror forms images purely by reflection. Its focal length is

$f = \frac{R}{2},$ …

Start your 14-day free trial to see the full solution →

Q.(a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope.

Part (a): Astronomical Telescope and Resolving Power

🎓Unlock everything free for 14 days

Unlock everything free for 14 days