Question

An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?

I tried to use lens maker's formula. As per me, image distance should increase. Can anyone confirm?

Answer 1

The image distance will be increased

Explanation:

We are to use here the Lens maker's for as stated below

`1/f=(mu_2/mu_1-1)(1/r_1-1/r_2)`

Where f =focal lens of the lens
`mu_1=`refractive index of the surrounding medium
`mu_2=`refractive index of medium of the lens
`r_1=` radius of curvature of the lens of first refracting surface where the beam is incident.
`r_2=` radius of curvature of the lens of the second refracting surface through which emergent beam comes out.

Here `r_1`and `r_2` remaining same and normally `mu_2> mu_1`

If the refractive index of surrounding medium i.e. `mu_1` is increased but kept less than `mu_2` then the ratio `mu_2/ mu_1` will be diminished and as a result the focal length f will be increased.
we know
`1/v-1/u=1/f`
where v = image distance
u = object distance
object distance remaining same the image distance proportionately increases with the increase in focal length.