A lens of refractive index n is put in a liquid of refractive index `n'`. If focal length of lens in air is `f`, its focal length in liquid will be.

Options

A. `(-fmu'(mu-1))/(mu-mu)`
B. `(-f(mu-mu))/(mu(mu-1))`
C. `(mu'(mu-1))/(f(mu-mu))`
D. `f(mu'mu)/(mu-mu')`

Answer

A

Explanation

According to question
Here `mu=`refractive index of lens
`mu=`refractive index of liquid focal length of lensin air =f

we have lens maker's `formula(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(1)))`
When lens is depped in water.
`1/f'=(mu_(g))/(mu_(i))-1))((1)/(R_(1))-(1)/(r_(2))) or (mu_(g)-mu))/(mu_(i))/(mu_(i))((1)/(R_(1))-(1)/(R_(2)))`
`therefore f'/f=((mu-1)mu')/(mu-mu')=-((mu-1)mu')/(mu'-mu)`
or `f=(-fmu'(mu-1))/(mu'-mu)`