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?​

Answer 1

Answer:

We know that

⇒

f

1

=(

μ

1

μ

2

−1)(

r

1

1

−

r

2

1

)

where

⇒f= focal length

⇒μ

1

=Refractive index of surrounding medium

⇒μ

2

=Refractive index of the lens

⇒r

1

=Radius of curvature of lens of first refracting surface were beam is incident

⇒r

1

=Radius of curvature of lens of second refracting surface through which beam is emerging out

If refractive index of surrounding medium μ

1

increases, but is kept less tha μ

2

, the the ratio

μ

1

μ

2

is decreased. So focal length is increased

We know that

⇒

v

1

−

u

1

=

f

1

Hence, object distace remains same,but image distance increaseswith increase in focal length.