Question

An equiconvex lens of refractive index 1.5 and focal length 10 cm (in air) is placed on the surface of water $(\mu =\frac{4}{3})$ such that its lower surface is immersed in water but its upper surface is in contact with air outside. Principal axis is perpendicular to water. A beam parallel to its principal axis is incident on lens from air then

• A. The beam will focus at a distance 20 cm from lens
• B. The beam will focus at a distance 30 cm from lens
  
• C. Focal length if lens wholly immersed in water will be 40 cm
  
• D. Focal length of lens wholly immersed in water will be 30 cm

Answer 1

Answer: (A), (C)

The correct options are
A The beam will focus at a distance 20 cm from lens
C Focal length if lens wholly immersed in water will be 40 cm



(a) $\frac{1}{f}=\frac{1.5-1}{1}\left(\frac{2}{R}\right)\Rightarrow f=R=10cm$

$\frac{\frac{4}{3}}{v}-\frac{1}{-\infty }=\frac{\frac{3}{2}-1}{10}+\frac{\frac{4}{3}-\frac{3}{2}}{-10}$

$\Rightarrow v=20cm$

(B) $\frac{1}{f}=\frac{1.5-\frac{4}{3}}{\frac{4}{3}}\times \frac{2}{10}\Rightarrow f=40cm$