Question

A 10cm thick rectangular glass slab is placed over a point object. A beaker
water up to a height of 10 cm is placed over the slab.
(a) Find the apparent location of the object when viewed normally.
(b) On moving the eye slowly away from the normal, the object disappears at a certain
position of the eye. (Given - refractive index of glass - 3/2 and water - 4/3)​

Answer 1

• Answer:

rectangular glass block of thickness 10 cm and refractive index 1.5 placed over a small coin. A beaker is filled with water of refractive index 4/3 to a height of 10 cm and is placed over the glass block.

(a) Find the apparent position of the object when it is viewed at near normal incidence.

(b) if the eye is slowly moves away from the normal at a certain position, the object is found to disappear, due to total internal reflection. At what surface does this happen and why ?

check-circle

Text Solution

Solution :

(a) Total shift

OI=t1(1−1/μ1)+t2(1−1/μ2)

=10(1−13/2)+10(1−14/3)

=103+104

Distance of image from P

PI=PO−OI=(10+10)−(103+104)

=14.17cm

(b) The image of coin disappears when the ray suffers T.I.R. at either at AB or CD.

The critical angle for galss-water surface

μ1sinC1=μ2sin90∘

32sinC1=43

sinC1=89

C1=sin−1(8/9)=sin−1(0.88)

The critical angle for water-air surface

μ2sinC2=1×sin90∘

43sinC2=1

sinC2=34

C2=sin−1(0.75)

C2>C1, T.I.R. occurs earlier at the water-air surface as eye is moved away from the normal.

Related Videos

VIEW ALL

An air bubble in a glass slab with refractive index `1.5` (near normal incidence) is `5 cm` deep when viewed from one surface and `3cm` deep when viewed from the opposite face. The thickness (in `cm)` of the slab is