Question

What happens to the focal length of a converging lens when it is placed under water?

Answer 1

Step-by-step explanation:

According to the lens maker's formula

The focal length of the converging lens in air is given by

`(1)/(f_(a))=\left ( \mu _(a)-1 \right )\left ( (1)/(R_(1))-(1)/(R_(2)) \right )` .... (1)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of glass in air.

When the lens is placed in water. the focal length is given by

`(1)/(f_(w))=\left ( (\mu _(a))/(\mu _(w))-1 \right )\left ( (1)/(R_(1))-(1)/(R_(2)) \right )` .... (2)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of lens in air and μw is the refractive index of glass in water.

Dividing equation (1) by (2), we observe that the value of focal length,

the focal length of lens in water increases.

Answer 2

When a converging lens is placed under water then its focal length increases.

Step-by-step explanation:

When any lens, be it converging or diverging is immersed into water then the speed of light before and after the refraction through the lens has less speed and due to less difference in speed it shows less deviations as a result the focal length is increased for the lenses.

Lesser the difference between the refractive index of the medium and the lens glass greater becomes its focal length.

The focal length of the lens immersed in a medium is given by:

`(1)/(f) =(n_l-n_m)/(n_m) * ((1)/(R_1) +(1)/(R_2) )`

where:

`n_l=` refractive index of the glass material with resp. to air

`n_m=` refractive index of the medium with resp. to air

`R_1\ \&\ R_2=` radii of curvature of the two surfaces of the lens