Question

A plano convex glass lens ($\mu_g$=3/2) of radius of curvature R = 10 cm is placed at a distance of 'b' from a concave lens of focal length 20 cm. The distance 'a' of a point object O from the plano convex lens so that the distance of final image from concave lens is independent of 'b' (in cm) is:

Answer 1

20

Answer 2

To solve this problem, we need to determine the conditions under which the final image formed by the two-lens system is independent of the distance 'b' between the lenses.

1. Calculate the focal length of the plano-convex lens (Lens 1): The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

For a plano-convex lens, one surface is plane (radius of curvature $R_2 = \infty$) and the other is convex (radius of curvature $R_1 = +10$ cm, assuming the convex surface faces the object). Given $\mu_g = 3/2$ and $R = 10$ cm.

$\frac{1}{f_1} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{+10} - \frac{1}{\infty} \right)$ $\frac{1}{f_1} = \left( \frac{1}{2} \right) \left( \frac{1}{10} - 0 \right)$ $\frac{1}{f_1} = \frac{1}{20}$

So, the focal length of the plano-convex lens is $f_1 = +20$ cm.

2. Analyze the image formation by the two-lens system: Let $u_1$ be the object distance for the first lens (plano-convex) and $v_1$ be the image distance. Let $u_2$ be the object distance for the second lens (concave) and $v_2$ be the final image distance. The object O is placed at a distance 'a' from the plano-convex lens, so $u_1 = -a$. The focal length of the concave lens is $f_2 = -20$ cm.

The condition that "the distance of the final image from the concave lens is independent of 'b'" implies that the rays incident on the concave lens must be parallel to the principal axis. If the rays incident on the concave lens are parallel, then the image formed by the first lens ($I_1$) must be at infinity. For a convex lens, an image is formed at infinity when the object is placed at its principal focus. Therefore, for the image $I_1$ formed by the plano-convex lens to be at infinity ($v_1 = \infty$), the object O must be placed at the focal point of the plano-convex lens. So, the distance 'a' must be equal to the focal length $f_1$.

$a = f_1 = 20 \text{ cm}$

3. Verify the condition: If $a = 20$ cm, then the object O is at the focal point of the plano-convex lens. According to the lens formula for Lens 1:

$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$ $\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{20}$ $\frac{1}{v_1} + \frac{1}{20} = \frac{1}{20}$ $\frac{1}{v_1} = 0 \implies v_1 = \infty$

So, the image $I_1$ is formed at infinity. This means the rays emerging from the plano-convex lens are parallel.

These parallel rays then act as the object for the concave lens. When parallel rays are incident on a lens, the image is formed at its focal point. For Lens 2 (concave lens), the object distance is $u_2 = \infty$. Using the lens formula for Lens 2:

$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$ $\frac{1}{v_2} - \frac{1}{\infty} = \frac{1}{-20}$ $\frac{1}{v_2} = \frac{1}{-20}$ $v_2 = -20 \text{ cm}$

The distance of the final image from the concave lens is $|v_2| = 20$ cm. This value is constant and does not depend on 'b'.

Therefore, the distance 'a' must be 20 cm.