Ray Optics Question 3
Question 3 - 24 January - Shift 2
A convex lens of refractive index 1.5 and focal length $18 cm$ in air is immersed in water. The change in focal length of the lens will be $cm$.
(Given refractive index of water $=\frac{4}{3}$ )
Show Answer
Answer: (54)
Solution:
Formula: Apparent Depth
$ \begin{aligned} & \frac{I}{f _{H_2 O}^{n}}=(\frac{\mu_g}{\mu _{H_2 O}}-1)(\frac{2}{R}) \\ & =\frac{1}{8}(\frac{2}{R}) \\ & =\frac{1}{(4 f _{\text{air }})} \end{aligned} $
So, $f _{H_2 O}=4 f _{\text{air }}=72 cm$
So change in focal length $=72-18=54 cm$
