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 3.5R
• 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

Answer: (A)

$R_1=-R,R_2+R,{\mu }_g=1.5and{\mu }_m=1.75$
$\therefore \frac{1}{f}=(\frac{{\mu }_g}{{\mu }_m}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Substituting the values, we have
$\frac{1}{f}=(\frac{1.5}{1.75}-1)(\frac{1}{-R}-\frac{1}{R})=\frac{1}{3.5R}$
$\therefore f=+3.5R$
Therefore, in the medium it will behave like a convergent
lens of focal length 3.5R. It can be understood as, ${\mu }_m>{\mu }_g,$
the lens will change its behaviour.