Question

A thin convergent glass lens $\mu =$1.5 has a power of $+5D$. When this lens is immersed in a liquid of refractive index $\mu$ it acts as a diverging lens of focal length 100 cm. The value of $\mu$ must be :

• A. $5/3$
• B. $4/3$
• C. $5/4$
• D. $6/5$

Answer 1

Answer: (A)

Given,

For convergent lens,

${\mu }_a=1$, refractive index of air

${\mu }_g=1.5$, refractive index of glass

$P_C=+5D$

the Power of the lens is given by the lens maker formula

$P_C=\frac{1}{f}=(\frac{{\mu }_g}{{\mu }_a}-1)(\frac{1}{R_1}-\frac{1}{R_2})=+5D$. . . . . .(1)

Now, When convergent lens immersed in a liquid of refractive index $\mu$, convergent lens act as a diverging lens of focal length $-100cm$

By lens maker formula,

$P_d=\frac{1}{f}=(\frac{{\mu }_g}{\mu }-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{-100}{100}=-1D$. . . . . . (2)

Divide equation (1) by equation (2), we get

$\frac{P_C}{P_d}=\frac{(\frac{{\mu }_g}{{\mu }_a}-1)}{\frac{{\mu }_g}{\mu }-1}=-5$

$\frac{{\mu }_g}{{\mu }_1}=-5(\frac{{\mu }_g}{\mu }-1)$

$\frac{1.5}{1}-1=-5(\frac{1.5}{\mu }-1)$

$\frac{0.5}{5}=-(\frac{1.5}{\mu }-1)$

$\mu =\frac{5}{3}$