A thin lens of refractive index `1.5` has focal length of `15 cm` in air. When the lens is placed is a medium of refractive index (4)/(3), its focal length will become …..cm.

Explanation

Accoding to lensmaker' formula
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))` (i)
Given refraction index of the lens,i.e.,
`._(S)^(a)mu=1.5=(mu_(g))/(mu_(a))`
Also, given refractive index of medium
`._(m)^(a)mu=(4)/(3)(mu_(m))/(mu_(a))`
`:. . _(g)^(m)mu=(mu_(g))/(mu_(m))=(mu_(g))/(mu_(a))xx(mu_(a))/(mu_(m))=(1.5)/(4//3)=1.125`
Applying Eq. (i) for th etwo cases, we get
`(1)/(15)=(1.5-1)((1)/(R_(1))-(1)/(R_(2)))`
and `(1)/(f_(2))=(1.5-1)/(1.125-1)=((1)/(R_(1))-(1)/(R_(2)))`
On dividing, we get
`(f_(2))/(15)=(1.5-1)/(1.125-1)=(0.5)/(0.125)=4rArrf_(2)=60cm`