Question:
A convex lens of refractive index
\[
\frac{3}{2}
\]
has a power of
\[
5\,D.
\]
If it is placed in a liquid of refractive index \(2\), the new power of the lens is
A convex lens of refractive index
\[
\frac{3}{2}
\]
has a power of
\[
5\,D.
\]
If it is placed in a liquid of refractive index \(2\), the new power of the lens is
Show Hint
For a lens immersed in a medium,
\[
P_{\text{medium}}
=
\frac{\left(\frac{n_l}{n_m}-1\right)}
{(n_l-1)}
\,P_{\text{air}}.
\]
If the medium has a higher refractive index than the lens, the power becomes negative and a convex lens behaves like a concave lens.
Updated On: Jul 29, 2026
- \(2.5\,D\)
- \(1.25\,D\)
- \(-1.25\,D\)
- \(-2.5\,D\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
The power of a lens in a medium is given by
\[
P=
\left(
\frac{n_l}{n_m}-1
\right)
\left(
\frac{1}{R_1}-\frac{1}{R_2}
\right),
\]
where
\[
n_l=\text{refractive index of lens},
\qquad
n_m=\text{refractive index of medium}.
\]
Step 1: Find the curvature factor of the lens. In air, \[ P_1 = (n_l-1) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] Given, \[ P_1=5D, \qquad n_l=\frac32. \] Hence, \[ 5 = \left( \frac32-1 \right) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] \[ 5 = \frac12 \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] Therefore, \[ \left( \frac{1}{R_1}-\frac{1}{R_2} \right) = 10. \]
Step 2: Calculate the new power in the liquid. Given, \[ n_m=2. \] Thus, \[ P_2 = \left( \frac{\frac32}{2}-1 \right) (10). \] \[ = \left( \frac34-1 \right) (10). \] \[ = \left( -\frac14 \right) (10). \] \[ =-2.5D. \]
Step 3: Interpret the result. The power becomes negative because the refractive index of the surrounding liquid is greater than that of the lens. Hence the convex lens behaves like a diverging lens. Therefore, \[ \boxed{P=-2.5D} \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: Find the curvature factor of the lens. In air, \[ P_1 = (n_l-1) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] Given, \[ P_1=5D, \qquad n_l=\frac32. \] Hence, \[ 5 = \left( \frac32-1 \right) \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] \[ 5 = \frac12 \left( \frac{1}{R_1}-\frac{1}{R_2} \right). \] Therefore, \[ \left( \frac{1}{R_1}-\frac{1}{R_2} \right) = 10. \]
Step 2: Calculate the new power in the liquid. Given, \[ n_m=2. \] Thus, \[ P_2 = \left( \frac{\frac32}{2}-1 \right) (10). \] \[ = \left( \frac34-1 \right) (10). \] \[ = \left( -\frac14 \right) (10). \] \[ =-2.5D. \]
Step 3: Interpret the result. The power becomes negative because the refractive index of the surrounding liquid is greater than that of the lens. Hence the convex lens behaves like a diverging lens. Therefore, \[ \boxed{P=-2.5D} \] \[ \boxed{\text{Answer = (D)}} \]
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