CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A point object O is placed in front of a transparent slab at a distance x from its closer surface. It is seen from the other side of the slab by light incident nearly normally to the slab. The thickness of the slab is t and its refractive index is μ. The apparent shift in the position of the object is


A

t(1+tμ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

t(1tμ)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

t(2tμ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

t(2+tμ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

t(1tμ)



The situation is shown in figure. Because of the refraction at the first surface, the image of O is formed at O1. For this refractiion, the real depth is AO = x and the apparent depth is AO1. Also the first medium is air and the second is the slab. Thus,
xAO1=1μor,AO1=μx.
The point O1 acts as the object for the refractiion at the second surface. Due to this refraction the image of O1 is formed at I. Thus,
BO1BI=μ
or, AB+AO1BI=μ or t+μxBI=μ
or BI=x+tμ.
The net shift in OI=OBBI=(x+t)(x+tμ)
=t(1tμ)
which is independent of x.


flag
Suggest Corrections
thumbs-up
4
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Playing with Glass Slabs
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon