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# 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:-

A
$$0.4 u$$
B
$$0.6 u$$
C
$$0.8 u$$
D
$$0$$
Solution
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#### Correct option is D. $$0$$$$\begin{array}{l}\text { calculating focus in first case }\rightarrow\\R_{1}=10\mathrm{~cm}\\R_{2}=5\mathrm{~cm} \\\mu=\text { refractive index }\end{array}$$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 lensdoen't change by reversingits orientation.So the difference in image position=0Hence option ' $$D$$ is correct.

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