Question

A convex lens made of glass $({\mu }_g=3/2)$ has focal length $f$ in air. The image of an object placed infront of it is inverted real and magnified. Now the whole arrangement is immersed in water $({\mu }_w=4/3)$ without changing the distance between object and lens. Then

• A. The new focal length will become $4f$
• B. The new focal length will become $f/4$
• C. New image will be virtual and magnified
• D. New image will be real inverted and smaller in size.

Answer 1

Answer: (A), (C)

The correct options are
A The new focal length will become $4f$
C New image will be virtual and magnified

We know,

Lens maker formula,

$\frac{1}{f}=(\frac{{\mu }_L}{{\mu }_M}-1)(\frac{1}{R}-\frac{-1}{R})$

When lens was in air,

$\frac{1}{f_1}=(\frac{3}{2}-1)\frac{2}{R}=\frac{1}{R}$

When lens was in medium of $\mu =\frac{4}{3}$

$\frac{1}{f_2}=(\frac{\frac{3}{2}}{\frac{4}{3}}-1)\frac{2}{R}=\frac{1}{4R}$

$f_1=R,f_2=4R$

In case 1 , object was between f and 2f,

But in case 2 the same object will be inside the focus and pole, as $f_2=4f$

Hence the image in case 2 will be virtual and magnified.

Option $\text{A,C}$ is correct answer.