Question

A convex lens refractive index is 1.5 has a power p if it is immersed in a liquid ( refractive index is 4/3 ) then its power will

Answer 1

power of lens depends on focal length of lens. e.g., power is inverse of focal length.

and we know, focal length depends on refractive index of medium.

it is given by , $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

case 1 : refractive index of lens, $\mu$ = 1.5

so, $\frac{1}{f}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$P=\frac{1}{f}=(0.5)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$......(1)

case 2 : lens is immersed in a liquid of refractive index , $\mu_l=4/3$

so, new focal length of lens , f' is given by, $\frac{1}{f'}=\left(\frac{1.5}{4/3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

or, $P'=\frac{1}{f'}=(0.125)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$......(2)

from equations (1) and (2),

$\frac{P}{P'}=\frac{(0.5)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}{(0.125)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$

or, $\frac{P}{P'}=\frac{0.5}{0.125}=\frac{4}{1}$

or, $P'=\frac{P}{4}$

Answer 2

Answer:

The power is $\dfrac{P}{4}$.

Explanation:

Given that,

Refractive index of lens= 1.5

Refractive index of liquid= 4/3

Power :

Power is inverse of focal length.

$P=\dfrac{1}{f}$

We need to calculate the focal length of convex lens

Using formula of focal length

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

Put the value of refractive index of lens

$P=\dfrac{1}{f}=(1.5-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$...(I)

After immersed in liquid,

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

Put the value of refractive index of lens

$P'=\dfrac{1}{f'}=(\dfrac{1.5}{\dfrac{4}{3}}-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

$P'=\dfrac{1}{f'}=(1.125-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$...(II)

Divided equation (I) by (II)

$\dfrac{P}{P'}=\dfrac{(1.5-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})}{(1.125-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})}$

$\dfrac{P}{P'}=\dfrac{1.5-1}{1.125-1}$

$\dfrac{P}{P'}=\dfrac{4}{1}$

$P'=\dfrac{P}{4}$

Hence, The power is $\dfrac{P}{4}$.