Question

A thin lens made of a material of refractive index μ2 has a medium of refractive index μ1, on one side and a medium of refractive index μ3, on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium μ1, and (b) from the medium μ3 ?

Answer 1

A) Image formed if the beam is incident from (a) the medium μ1:

μ2/V-μ1/u=μ2-μ1/R

μ2/v-μ1/∞=μ2-μ1/R

1/V= μ2-μ1/Rμ2

v=Rμ2/μ2-μ1

For refraction at other side:

μ3/v - μ2/u= μ3-μ2/R

here image formed due to first refraction  becomes virtual object here

μ3/v= -[ μ3-μ2/R -μ2/u ]

μ3/v= -[μ3-μ2/R  - μ2(μ2-μ1)/Rμ2

μ3/v= -[ μ3-μ2-μ2+μ1]/R

μ3/v = - [μ3-2μ2+μ1]/R

1/v= - [μ3-2μ2+μ1]/Rμ3

v= -Rμ3/μ3-2μ2+μ1

Therefore Image is formed at v= Rμ3/2μ2-μ3-μ1

Similarly

image formed if the beam is incident from the medium μ3 :

V=μ1R/2μ2-μ1-μ3

Answer 2

Answer:

a) When the beam is incident on the lens from medium μ

1

.

Then

v

μ

2

−

u

μ

1

=

R

μ

2

−μ

1

or

v

μ

2

−

(−∞)

μ

1

=

R

μ

2

−μ

1

or

v

1

=

μ

2

R

μ

2

−μ

1

or v=

μ

2

−μ

1

μ

2

R

Again, for 2ns refraction,

v

μ

3

−

u

μ

2

=

R

μ

3

−μ

2

or,

v

μ

3

=−[

R

μ

3

−μ

2

−

μ

2

R

μ

2

(μ

2

−μ

1

)]

⇒−[

R

μ

3

−μ

2

−μ

2

+μ

1

]

or, v=−[

μ

3

−2μ

2

+μ

1

μ

3

R

]

So, the image will be formed at =

2μ

2

−μ

1

−μ

3

μ

3

R

b) Similarly for the beam from μ

3

medium the image is formed at

2μ

2

−μ

1

−μ

3

μ

1

R

.