Question

A concave lens of glass , refractive index 1.5, has both surfaces of same radius of curvature R.On immersion in a medium of refractive index 1.75, it will behave as a :-

• A. Convergent lens of focal length 3.5 R
• B. Convergent lens of focal length 3.0 R
• C. Divergent lens of focal length 3.5 R
• D. Divergent lens of focal length 3.0 R

Answer 1

The correct option is A Convergent lens of focal length 3.5 R

Using lens maker formula :
$\frac{1}{f}=(\frac{n_{lens}}{n_{surr}}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Given : $R_1=-R$ $R_2=R$ $n_{lens}=1.5$ $n_{surr}=1.75$
$\therefore$ $\frac{1}{f}=(\frac{1.5}{1.75}-1)(\frac{1}{-R}-\frac{1}{R})$
$⟹f=+3.5R$
Since the focal length of the lens immersed in medium is positive, thus the concave lens behaves as a convergent lens in that medium.