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Physics Question on Spherical Mirrors

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

convergent lens of focal length 3.5R

Explanation

Solution

The  correct  answer  is  A:convergent  lens  of  focal  length  3.5RThe \space correct \space answer\space is \space A:convergent\space lens\space of\space focal\space length\space 3.5R
R1=−R,R2+R,μg=1.5andμm=1.75R_1=-R,R_2+R, \mu_g=1.5 and \mu_m=1.75
∴          1f=(μgμm−1)(1R1−1R2)\therefore \, \, \, \, \, \, \, \, \, \, \frac{1}{f}=\Bigg(\frac{\mu_g}{\mu_m}-1\Bigg)\Bigg(\frac{1}{R_1}-\frac{1}{R_2}\Bigg)
Substituting the values, we have
1f=(1.51.75−1)(1−R−1R)=13.5R\frac{1}{f}=\Bigg(\frac{1.5}{1.75}-1\Bigg)\Bigg(\frac{1}{-R}-\frac{1}{R}\Bigg)=\frac{1}{3.5 R}
∴           f=+3.5R\therefore \, \, \, \, \, \, \, \, \, \, \, f=+3.5 R
Therefore, in the medium it will behave like a convergent
lens of focal length 3.5R. It can be understood as, μm>μg,\mu_m > \mu_g,
the lens will change its behaviour.