Solve

Guides

Question

A thin convex lens is made of a material of refractive index 1.6. An object is kept at a distance of a from lens on the principal axis as shown in the figure. The radius of curvature of the surface are 10 cm and 5 cm. Now, the lens is reversed such that the face having radius of curvature 5 cm lies close to the object. The difference in image position as obtained for both the cases is equal to:-

Open in App

Solution

Verified by Toppr

$$\begin{array}{l}\text { calculating focus in first case }\rightarrow\\R_{1}=10\mathrm{~cm}\\R_{2}=5\mathrm{~cm} \\\mu=\text { refractive index }\end{array}$$

using formula

$$\begin{aligned} \frac{1}{f} &=\left(\frac{\mu}{1}-4\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \frac{4}{f} &=(1.6-1)\left(\frac{1}{10}-\frac{1}{-5}\right) \\ \frac{1}{f} &=(0.6)\left[\frac{1}{10}+\frac{1}{5}\right) \\ f &=50 / 9 \mathrm{~cm} \end{aligned}$$

calculating focus in second case $$\rightarrow$$

$$\begin{array}{l}R_{1}=\operatorname{5cm} \\R_{2}=10\mathrm{~cm}\end{array}$$

$$\frac{1}{f}=(0.6)\left[\frac{1}{5}+\frac{1}{10}\right]$$

$$f=\frac{50}{9} \mathrm{~cm}$$

The focus of lens

doen't change by reversing

its orientation.

So the difference in image position=0

Hence option ' $$D$$ is correct.

Was this answer helpful?

0