Question

The focal length of lens of refractive index 2.5 in air is 30 cm. If it is immersed in a liquid of refractive index 53 , then its focal length will be​

Answer 1

Answer:

For the lens in air

fa1=(aμg−1)[R11−R21]

⇒151=(1.5−1)[R11−R21]

⇒152=R11−R21

When the lens is immersed in water

fw1=(aμwaμg−1)[R11−R21]

⇒fw

use this formula and solve question

Answer 2

Given: Refractive Index, $\mu_{air} = 2.5$

Focal Length, f =30cm,

Refractive index of liquid, $\mu_{liquid} = 53$

Find: Here we have to calculate the focal length with respect to the water,

Solution:

We have the expression,

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

$\dfrac{1}{30} = (2.5 - 1)(\dfrac{1}{R_{1} } - \dfrac{1}{R_{2} } )$

$\dfrac{1}{45} = (\dfrac{1}{R_{1} } - \dfrac{1}{R_{2} } )$

Refractive index of glass with respect to the glass,

$\mu_{avg} =\dfrac{\mu_{air} }{\mu_{liquid} } = \dfrac{2.5}{53} = 0.0471$

Now, we calculate the focal length,

$\dfrac{1}{f'} = (\mu_{avg} - 1)(\dfrac{1}{R_{1} } - \dfrac{1}{R_{2} } )$

$\dfrac{1}{f'} = (0.7001 - 1)(\dfrac{1}{R_{1} } - \dfrac{1}{R_{2} } )$

$\dfrac{1}{f'} = (-0.2999)(\dfrac{1}{45} })\\\\\dfrac{1}{f'} = -0.0066cm$

The value of the focal length, f = -0.0066cm