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

The Correct Option is A

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