Question

A double convex lens made of glass of refractive index \(1.5\) and radii of curvature \(20\,\text{cm}\) each is immersed in a liquid of refractive index \(n_L\). The correct plot showing the variation of the power, in the units of diopter \((D)\), as a function of \(n_L\), is:

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Power of a lens in a medium: \[ P= \left(\frac{n_g}{n_m}-1\right) \left(\frac1{R_1}-\frac1{R_2}\right) \] If surrounding medium has larger refractive index than the lens, the lens may behave like a concave lens.

Answer 1

The Correct Option is C

Step 1: Use lens maker formula in a medium.
Power of a lens in a medium: \[ P= \left(\frac{n_g}{n_L}-1\right) \left(\frac1{R_1}-\frac1{R_2}\right) \] Here: \[ n_g=1.5 \] For double convex lens: \[ R_1=+20\ \text{cm}=0.2\ \text{m} \] \[ R_2=-20\ \text{cm}=-0.2\ \text{m} \] Thus: \[ \left(\frac1{R_1}-\frac1{R_2}\right) = \frac1{0.2}-\left(-\frac1{0.2}\right) \] \[ =5+5 \] \[ =10 \] Hence: \[ P= 10\left(\frac{1.5}{n_L}-1\right) \]

Step 2: Analyze the nature of the graph.
Rewrite: \[ P= \frac{15}{n_L}-10 \] This is of the form: \[ y=\frac{a}{x}+b \] Hence graph is: \[ \text{hyperbolic} \]

Step 3: Find where power becomes zero.
Set: \[ P=0 \] \[ \frac{15}{n_L}-10=0 \] \[ \frac{15}{n_L}=10 \] \[ n_L=1.5 \] Thus graph crosses: \[ P=0 \] at: \[ n_L=1.5 \] For: \[ n_L<1.5 \] \[ P>0 \] For: \[ n_L>1.5 \] \[ P<0 \] This matches option: \[ \boxed{\mathrm{(C)}} \]

Step 4: Identify the correct option.
Therefore: \[ \boxed{\mathrm{(C)}} \]