Question

What will happen to the focal length of a convex lens of refractive index 4/3,when placed in water of refractive index 3/2.give reasons.

Answer 1

its focal lenght will change as medium change firstly the incident light travels from air to lens
but when it is in water light has to travel through water and air then it enter into the lens..

Answer 2

Answer: The focal length of the lens in water is reduced to two by third of focal length of the lens in air.

Explanation:

Given that,

Refractive index $\mu=\dfrac{4}{3}$

Refractive index of water $\mu=\dfrac{3}{2}$

We know that,  

The focal length of the lens is inversely proportional to the refractive index.

When the refractive index will be decreases then the focal length will be increases.

$\dfrac{1}{f}= (\mu-1)(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

The focal length of the lens in the air

$\dfrac{1}{f_{a}}= (\mu_{a}-1)(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$....(I)

The focal length of the lens in the water

$\dfrac{1}{f_{w}}= (\mu_{w}-1)(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$....(II)

Divided equation (I) by equation (II)

$\dfrac{f_{w}}{f_{a}}=\dfrac{\mu_{a}-1}{\mu_{w}-1}$

$\dfrac{f_{w}}{f_{a}}=\dfrac{\dfrac{4}{3}-1}{\dfrac{3}{2}-1}$

$\dfrac{f_{w}}{f_{a}}=\dfrac{2}{3}$

Hence, The focal length of the lens in water is reduced to two by third of focal length of the lens in air.