Question

The two surface of a concave lens, made of glass of refractive index $1.5$ have the same radii of curvature $R$. It is now immersed in a refractive index $1.75$, then the lens:

• A. becomes a convergent lens of focal length $3.5R$
• B. becomes a convergent lens of focal length $3.0R$
• C. changes as a divergent lens of focal length $3.5R$
• D. changes as a divergent lens of focal length $3.0R$

Answer 1

Answer: (A)

becomes a convergent lens of focal length $3.5R$

$n_{21}=\frac{n_2}{n_1}=\frac{1.5}{1.75}=\frac{6}{7}$

From lens maker's formula,

$\frac{1}{f}=(n_{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f}=(\frac{6}{7}-1)(\frac{1}{-R}-\frac{1}{R})$

Solving, $f=3.5R$

Focal length is positive. Hence, lens is convergent.