Question

There is a spherical glass shell of refractive index 1.5, inner radius 10 cm and outer radius 20 cm. Inside the spherical cavity, there is air. A point object is placed at a point O at a distance of 30cm from the outer spherical surface. Find the final position of the image as seen by eye

• A. 11.75 cm from centre
  
• B. 28.75 cm from centre
  
• C. 78.75 cm from centre
  
• D. 18.75 cm from centre
  

Answer 1

The correct option is D 18.75 cm from centre

Given: There is a spherical glass shell of refractive index 1.5, inner radius 10 cm and outer radius 20 cm. Inside the spherical cavity, there is air. A point object is placed at a point O at a distance of 30cm from the outer spherical surface

To find the final position of the image as seen by eye.

Solution:

For first surface,

$\frac{1.5}{\nu }-\frac{1}{-30}=\frac{1.5-1}{20}⟹\nu =-180cm$

For the second surface, $u_1=-190cm$

$\frac{1}{\nu }-\frac{1.5}{-190}=\frac{1-1.5}{10}⟹\nu =-\frac{190}{11}cm$

For the third surface, $u=-(\frac{190}{11}+20)=\frac{-410}{11}cm$

$\frac{1.5}{\nu }-\frac{11}{-410}=\frac{1.5-1}{-10}=-\frac{1}{20}⟹\nu =-\frac{-410}{21}cm$

For the fourth surface, $u=-(\frac{410}{21}+10)=\frac{-620}{21}cm$

${\mu }_1=1.5,{\mu }_2=1,R=-20cm$

$⟹\frac{1}{\nu }-\frac{1.5\times 21}{-620}=\frac{1-1.5}{-20}⟹\nu =-38.75cm$

The final image is at a distance of 18.75 cm from the center