Q.A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?

Concept and intuition

When an object and a screen are a fixed distance $D$ apart, a convex lens can form a sharp image on the screen in two distinct positions (provided $D > 4f$). At one position the image is magnified; at the other it is diminished. The distance between these two lens positions is $d$. This is the displacement method for measuring focal length. The key insight: the lens formula and symmetry imply that the two object distances are $u$ and $D - u$, and the difference between them is $d$. Solving the lens equation yields a neat formula that avoids solving for $u$ directly.

Step-by-step solution

1. Set up the variables

   Let the fixed distance between object and screen be $D = 90\ \text{cm}$.

   Let the distance between the two lens positions be $d = 30\ \text{cm}$.

   For the first position, let the object distance be $u$; then the image distance is $v = D - u$.

   For the second position, the roles swap: the object distance becomes $v$ and the image distance becomes $u$.

   The separation between the two lens positions is $|u - v| = d$.

2. Relate $u$ and $v$ to $D$ and $d$

   Since $u + v = D$ and $|u - v| = d$, we can solve:

(or vice versa; the absolute difference is $d$).

Here:

3. Apply the lens formula The thin lens equation:

Substitute $u = 60\ \text{cm}$, $v = 30\ \text{cm}$:

Hence $f = 20\ \text{cm}$.

4. Alternative direct formula The displacement method gives the general result:

Plugging in:

This confirms the calculation.

Concept & Intuition

In a compound microscope, the objective lens forms a real, inverted, and magnified image of the object. This image then acts as the object for the eyepiece (the second lens). When the final image is at infinity, the eyepiece is being used as a simple magnifier in its “normal adjustment” — meaning the intermediate image must be placed exactly at the eyepiece’s focal point. That gives us the distance from the eyepiece to the intermediate image. Since we know the separation between the two lenses, we can then find the distance from the objective to that intermediate image. Finally, using the lens formula for the objective, we solve for the object distance.

Step-by-step solution

1. Understand the setup

   • Objective focal length: $f_o = 2\ \text{cm}$
   • Eyepiece focal length: $f_e = 3\ \text{cm}$
   • Distance between objective and eyepiece (tube length, but not exactly the standard “tube length” definition here): $L = 15\ \text{cm}$
   • Final image is at infinity → rays entering the eyepiece are parallel → the intermediate image must lie at the first focal point of the eyepiece.

2. Locate the intermediate image

   For the eyepiece, if the object (intermediate image) is at its focal point, the image is at infinity. So the intermediate image is at a distance $f_e = 3\ \text{cm}$ in front of the eyepiece.

   Since the lenses are 15 cm apart, the distance from the objective to this intermediate image is:

This is the image distance for the objective lens.

3. Apply the lens formula to the objective The lens formula:

where $u_o$ is the object distance from the objective (positive if real object is on the same side as incoming light, but we treat distances as positive in the formula with sign conventions; here we use the real-is-positive convention for simplicity).

Given $f_o = 2\ \text{cm}$ and $v_o = 12\ \text{cm}$:

4. Interpret the result The object is placed 2.4 cm in front of the objective lens. The image formed by the objective is 12 cm behind it (real image). These match the pair (2.4 cm, 12 cm).

Concept and intuition

A person whose least distance of distinct vision (near point) is 50 cm cannot see objects clearly closer than that. To read at the normal near point of 25 cm, we need a lens that makes an object at 25 cm appear to be at 50 cm (the person’s actual near point). This is a classic case of a converging lens used as a simple magnifier for a hypermetropic (farsighted) eye. The lens creates a virtual image at the person’s near point, which the eye can then focus on.

Step-by-step reasoning

1. Identify the object and image distances

   The object is placed at the normal near point: $u = -25$ cm (negative by sign convention: object is on the same side as incoming light).

   The image must be formed at the person’s actual near point: $v = -50$ cm (negative because the image is virtual and on the same side as the object).

2. Apply the lens formula

   The thin lens equation is

Substituting the values (all in cm):

So the focal length $f = +50$ cm. The positive sign confirms a converging lens.

3. Convert focal length to power Power in dioptres is $P = \frac{1}{f(\text{in metres})}$.

4. Match with the options The power is $+2$ dioptres, which corresponds to option (B).

Concept and Intuition

This problem uses a common experimental technique called the displacement method (or conjugate foci method) to determine the focal length of a convex lens. The core idea is that for a fixed distance $D$ between an object and a screen, a convex lens can form a sharp, real image on the screen at two different positions, provided $D$ is greater than or equal to four times the focal length ($D \ge 4f$).

This phenomenon arises from the principle of conjugate foci. If an object at position $O$ forms an image at position $I$ through a lens, then if the object is placed at $I$, its image will be formed at $O$. In the displacement method, the object and screen are fixed. When the lens is at the first position ($L_1$), it forms an image of the object on the screen. If we then move the lens to a second position ($L_2$), another sharp image is formed on the screen. The key insight is that the object distance for the first position becomes the image distance for the second position, and vice-versa.

Let $u_1$ be the object distance and $v_1$ be the image distance for the first lens position.

Then, for the second lens position, the object distance will be $v_1$ and the image distance will be $u_1$.

The total distance between the object and the screen is $D = u_1 + v_1$.

The distance the lens is displaced between the two positions is $x = |v_1 - u_1|$.

These two equations allow us to find $u_1$ and $v_1$ in terms of $D$ and $x$, which can then be substituted into the lens formula to find the focal length $f$.

Step-by-step Solution

1. Identify the given values:

   • Distance between the object and the screen, $D = 90\ \text{cm}$.
   • Distance between the two lens positions, $x = 20\ \text{cm}$.

2. Apply the formula for focal length:

   The focal length $f$ using the displacement method is given by the formula:

3. Substitute the values and calculate: Substitute $D = 90\ \text{cm}$ and $x = 20\ \text{cm}$ into the formula:

4. Round to the nearest option: Rounding the calculated value to two decimal places, we get $f \approx 21.39\ \text{cm}$. Comparing this with the given options, $21.38\ \text{cm}$ is the closest approximation.

Concept and intuition

A symmetrical biconvex lens with equal radii means the two surfaces are identical in curvature. The lens maker’s formula gives the focal length directly from the radii and refractive index. Once we know the focal length, the thin lens equation tells us where the image forms relative to the optical center. But the question asks for the distance from the rear surface, not from the center. So we must account for the lens thickness — here the lens is thin enough that we can treat its center as the midpoint, but the problem expects us to realize the image distance from the rear surface is simply the image distance from the center minus half the (negligible) thickness? Actually, careful: For a thin lens, the distance from the rear surface is essentially the same as from the optical center, because the lens thickness is small compared to other distances. However, the problem gives radii of 5 cm, so the lens is not extremely thin — but in standard optics problems, “thin lens” approximation is used unless thickness is given. Here no thickness is given, so we assume the lens is thin and the optical center coincides with the lens center. Then the image distance from the rear surface equals the image distance from the optical center.

Let’s work it through.

1. Find the focal length using the lens maker’s formula For a lens in air (refractive index of air = 1), the formula is

For a biconvex lens, the first surface (front) has $R_1 = +5$ cm (positive because convex toward the object), and the second surface (rear) has $R_2 = -5$ cm (negative because convex toward the object means the center of curvature is on the opposite side).

So

Thus $f = 5$ cm.

2. Apply the thin lens equation The object is placed 20 cm from the front surface. For a thin lens, the object distance $u$ is measured from the optical center. Since the lens is thin, the front surface is essentially at the optical center, so $u = -20$ cm (negative by sign convention: object on the incident side). The lens equation:

Substituting:

So $v = \frac{20}{3}$ cm.

The positive sign means the image is real and on the opposite side of the lens from the object.

3. Distance from the rear surface Since the lens is thin, the rear surface is at the same location as the optical center. Therefore the image distance from the rear surface is simply $v = \frac{20}{3}$ cm.

The key to this problem is the mirror equation and the behaviour of rays from a concave mirror. When you place an object between the focal point $F$ and the centre of curvature $C$, the reflected rays converge to form a real image on the same side as the object. The exact position and nature of that image follow directly from the mirror formula and a quick ray diagram.

Let’s walk through it.

1. Recall the mirror equation For a spherical mirror,

where $f$ is the focal length (negative for concave), $u$ is the object distance (negative, by sign convention), and $v$ is the image distance (negative for real images).

The radius of curvature $R = 2f$, so $C$ is at distance $2f$ from the mirror.

2. Set up the object position The object lies between $F$ and $C$, so

(Remember: $u$ is negative, but we work with magnitudes for clarity.)

3. Find the image distance From the mirror equation:

Since $|u| > f$, the term $1/|u|$ is smaller than $1/|f|$, so $1/v$ is negative — meaning $v$ is negative, hence a real image.

Moreover, because $|u| < 2f$, we have $1/|u| > 1/(2f)$, so

This gives $|v| > 2|f|$, i.e. the image lies beyond the centre of curvature.

4. Determine magnification and orientation

   Magnification $m = -v/u$. Since both $v$ and $u$ are negative, $m$ is negative — meaning the image is inverted relative to the object.

   Also $|v| > |u|$, so $|m| > 1$ — the image is enlarged.

5. Ray diagram confirmation

   A ray parallel to the axis reflects through $F$; a ray through $F$ reflects parallel. For an object between $F$ and $C$, these two rays converge on the far side of $C$, forming an inverted, real, enlarged image.