Question

A plano-convex lens acts as a concave mirror of focal length 20 cm if it is silvered on the plane side. When the convex side is silvered it acts like a concave mirror of focal length 7 cm. The refractive index of the material of the lens is:

• A. $\frac{20}{13}$
• B. $\frac{7}{13}$
• C. $\frac{12}{7}$
• D. $\frac{15}{7}$

Answer 1

Answer: (A)

$\frac{20}{13}$

Answer 2

The problem involves a plano-convex lens silvered on two different sides, acting as a concave mirror. We need to find the refractive index ($\mu$) of the lens material. Let $R$ be the radius of curvature of the convex surface of the lens.

The focal length of a plano-convex lens is given by the lens maker's formula:

$\frac{1}{f_L} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

For a plano-convex lens, one surface is plane ($R_1 = R$, $R_2 = \infty$ or vice versa). Assuming light enters the convex surface first, $R_1=R$ and $R_2=\infty$.

So, $\frac{1}{f_L} = (\mu - 1) \frac{1}{R}$

The effective focal length $F_{eq}$ of a silvered lens system is given by the formula for combination of powers:

$\frac{1}{F_{eq}} = \frac{2}{f_L} + \frac{1}{f_M}$

where $f_L$ is the focal length of the lens and $f_M$ is the focal length of the silvered surface acting as a mirror.

Case 1: Plane side is silvered.

When the plane side is silvered, the light passes through the convex surface, reflects off the plane mirror, and passes through the convex surface again. The mirror is a plane mirror, so its focal length $f_M = \infty$. The effective focal length $F_{eq1} = 20$ cm (given as a concave mirror, so we use the magnitude). Using the formula:

$\frac{1}{F_{eq1}} = \frac{2}{f_L} + \frac{1}{\infty}$

$\frac{1}{20} = \frac{2}{f_L}$

$f_L = 40$ cm

Now, substitute $f_L$ into the lens maker's formula:

$\frac{1}{40} = (\mu - 1) \frac{1}{R}$

$R = 40(\mu - 1)$ (Equation 1)

Case 2: Convex side is silvered.

When the convex side is silvered, the light passes through the plane surface, reflects off the silvered convex surface, and passes through the plane surface again. The silvered convex surface acts as a concave mirror (as stated in the problem). Its radius of curvature is $R$. The focal length of a spherical mirror is $f_M = R/2$. The effective focal length $F_{eq2} = 7$ cm. Using the formula:

$\frac{1}{F_{eq2}} = \frac{2}{f_L} + \frac{1}{f_M}$

$\frac{1}{7} = \frac{2}{f_L} + \frac{1}{R/2}$

$\frac{1}{7} = \frac{2}{f_L} + \frac{2}{R}$

Substitute $f_L = 40$ cm (from Case 1):

$\frac{1}{7} = \frac{2}{40} + \frac{2}{R}$

$\frac{1}{7} = \frac{1}{20} + \frac{2}{R}$

Now, solve for $R$:

$\frac{2}{R} = \frac{1}{7} - \frac{1}{20}$

$\frac{2}{R} = \frac{20 - 7}{140}$

$\frac{2}{R} = \frac{13}{140}$

$R = \frac{2 \times 140}{13} = \frac{280}{13}$ cm (Equation 2)

Calculate the refractive index ($\mu$):

Equate Equation 1 and Equation 2:

$40(\mu - 1) = \frac{280}{13}$

$\mu - 1 = \frac{280}{13 \times 40}$

$\mu - 1 = \frac{7}{13}$

$\mu = 1 + \frac{7}{13}$

$\mu = \frac{13 + 7}{13}$

$\mu = \frac{20}{13}$

The refractive index of the material of the lens is $\frac{20}{13}$.