Question

A convex lens in air produces a real image having the same size as object. When the object and the lens are immersed in a liquid, the real image formed is enlarged three times the object size. Find the refractive index of the liquid. ($\mu_{glass}=1.5$)

• A. $\frac{12}{11}$
• B. $\frac{4}{3}$
• C. $\frac{9}{8}$
• D. $\frac{6}{5}$

Answer 1

Answer: (C)

$\frac{9}{8}$

Answer 2

Here's how to solve this problem:

1. Analyze the lens in air:

When a convex lens in air produces a real image of the same size as the object, it implies that the object is placed at twice the focal length ($2F_a$).

• Magnification $M = -1$ (real and inverted image).
• Object distance $u = -2F_a$.
• Image distance $v = 2F_a$.
• The magnitude of the object distance is $|u| = 2F_a$.

The lens maker's formula for a lens in air is:

$\frac{1}{F_a} = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad (1)$

Where $F_a$ is the focal length in air, $\mu_g$ is the refractive index of the glass (lens material), and $R_1, R_2$ are the radii of curvature of the lens surfaces.

2. Analyze the lens in liquid:

When the object and the lens are immersed in a liquid, a real image is formed, enlarged three times the object size.

• Magnification $M' = -3$ (real and inverted image).

• Let the new focal length in the liquid be $F_l$.

• Let the object distance be $u'$ and the image distance be $v'$.

• From magnification formula: $M' = \frac{v'}{u'} = -3$. Since $u'$ is negative for a real object, $v'$ must be positive. So, $v' = -3u'$. (e.g., if $u' = -x'$, then $v' = 3x'$).

• Using the lens formula: $\frac{1}{v'} - \frac{1}{u'} = \frac{1}{F_l}$

  Substitute $v' = -3u'$:

  $\frac{1}{-3u'} - \frac{1}{u'} = \frac{1}{F_l}$

  $\frac{1 - 3}{-3u'} = \frac{1}{F_l}$

  $\frac{-2}{-3u'} = \frac{1}{F_l}$

  $\frac{2}{3u'} = \frac{1}{F_l}$

  So, $u' = \frac{2}{3}F_l$. (Note: Here $u'$ is used as the magnitude of object distance for simplicity in this step. If we stick to sign convention, $u' = - \frac{2}{3}F_l$ and $v' = 2F_l$).

  Let's re-do this using $u' = -x'$ and $v' = 3x'$ as in thought process.

  $\frac{1}{3x'} - \frac{1}{(-x')} = \frac{1}{F_l}$

  $\frac{1}{3x'} + \frac{1}{x'} = \frac{1}{F_l}$

  $\frac{1+3}{3x'} = \frac{1}{F_l}$

  $\frac{4}{3x'} = \frac{1}{F_l} \implies x' = \frac{4}{3}F_l$.

• The magnitude of the object distance is $|u'| = \frac{4}{3}F_l$.

It is generally assumed that the object position relative to the lens remains the same unless specified otherwise. Therefore, we assume the object distance is the same in both cases:

$|u| = |u'|$

$2F_a = \frac{4}{3}F_l$

$F_l = \frac{3}{2}F_a \quad (2)$

The lens maker's formula for a lens in liquid is:

$\frac{1}{F_l} = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad (3)$

Where $\mu_l$ is the refractive index of the liquid.

3. Calculate the refractive index of the liquid ($\mu_l$):

Divide equation (3) by equation (1):

$\frac{1/F_l}{1/F_a} = \frac{\left(\frac{\mu_g}{\mu_l} - 1\right)}{(\mu_g - 1)}$

$\frac{F_a}{F_l} = \frac{\left(\frac{\mu_g}{\mu_l} - 1\right)}{(\mu_g - 1)}$

Substitute $F_l = \frac{3}{2}F_a$ from (2) into this equation:

$\frac{F_a}{(3/2)F_a} = \frac{\left(\frac{\mu_g}{\mu_l} - 1\right)}{(\mu_g - 1)}$

$\frac{2}{3} = \frac{\left(\frac{1.5}{\mu_l} - 1\right)}{(\mu_g - 1)}$

Given $\mu_g = 1.5$.

$\frac{2}{3} = \frac{\left(\frac{1.5}{\mu_l} - 1\right)}{(1.5 - 1)}$

$\frac{2}{3} = \frac{\left(\frac{1.5}{\mu_l} - 1\right)}{0.5}$

Multiply both sides by 0.5:

$0.5 \times \frac{2}{3} = \frac{1.5}{\mu_l} - 1$

$\frac{1}{3} = \frac{1.5}{\mu_l} - 1$

Add 1 to both sides:

$\frac{1}{3} + 1 = \frac{1.5}{\mu_l}$

$\frac{4}{3} = \frac{1.5}{\mu_l}$

Now, solve for $\mu_l$:

$\mu_l = 1.5 \times \frac{3}{4}$

$\mu_l = \frac{3}{2} \times \frac{3}{4}$

$\mu_l = \frac{9}{8}$

The refractive index of the liquid is $\frac{9}{8}$.