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?

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?

1 Answer

Impact of this question

11038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-11T01:20:46

The image distance will be increased

Explanation:

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

$\frac{1}{f} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{r_{1}} - \frac{1}{r_{2}})$ $1/f=(mu_2/mu_1-1)(1/r_1-1/r_2)$

Where f =focal lens of the lens
$μ_{1} =$ $mu_1=$ refractive index of the surrounding medium
$μ_{2} =$ $mu_2=$ refractive index of medium of the lens
$r_{1} =$ $r_1=$ radius of curvature of the lens of first refracting surface where the beam is incident.
$r_{2} =$ $r_2=$ radius of curvature of the lens of the second refracting surface through which emergent beam comes out.

Here $r_{1}$ $r_1$ and $r_{2}$ $r_2$ remaining same and normally $μ_{2} > μ_{1}$ $mu_2> mu_1$

If the refractive index of surrounding medium i.e. $μ_{1}$ $mu_1$ is increased but kept less than $μ_{2}$ $mu_2$ then the ratio $\frac{μ_{2}}{μ_{1}}$ $mu_2/ mu_1$ will be diminished and as a result the focal length f will be increased.
we know
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $1/v-1/u=1/f$
where v = image distance
u = object distance
object distance remaining same the image distance proportionately increases with the increase in focal length.

Science

Math

Humanities

... and beyond

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?

1 Answer

Explanation:

Related questions

Impact of this question