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Physics Question on Ray optics and optical instruments

A thin glass (refractive index 1.5) lens has optical power of -5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be

A

+0.25D

B

+1D

C

+0.625D

D

-1D

Answer

+0.625D

Explanation

Solution

For a thin lens
P=1f=(μgμa−1)(1R1−1R2)P=\frac{1}{f}=\left(\frac{\mu_{g}}{\mu_{a}} -1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
P′=1f′=(μgμl−1)(1R1−1R2)P' =\frac{1}{f'}=\left(\frac{\mu_{g}}{\mu_{l}} -1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
P′P=(μgμl−1)(μgμa−1)=(1.51.6−1)(1.51−1)=−0.11.60.5\frac{P'}{P} =\frac{\left(\frac{\mu_{g}}{\mu_{l}}-1\right)}{\left(\frac{\mu_{g}}{\mu_{a}}-1\right)}=\frac{\left(\frac{1.5}{1.6}-1\right)}{\left(\frac{1.5}{1}-1\right)}=\frac{\frac{-0.1}{1.6}}{0.5}
⇒  P′=−1×216×1P=−18(−5)=+0.625 D\Rightarrow \:\:P' = \frac{-1 \times 2}{16 \times1} P = \frac{-1}{8}(-5) = +0.625 \,D