wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

k transparent slabs are arranged one over another. The refractive indices of the slabs are μ1, μ2, μ3, ... μk and the thicknesses are t1 t2, t3, ... tk. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.

Open in App
Solution

k number of transparent slabs are arranged one over the other.
Refractive indices of the slabs = μ1, μ2, μ3, ..., μk
Thickness of the slabs = t1, t2, t3,.., tk
Shift due to one slab:
t=1-1μt

For the combination of multiple slabs, the shift is given by,
t=1-1μ1t1+1-1μ2t2+...+1-1μktk ...(i)

Let μ be the refractive index of the combination of slabs.
The image is formed at the same place.
So, the shift will be:

Δt=1-1μ(t1+t2 ...+tk) ...(ii)

Equating (i) and (ii), we get:

1-1μ(t1+t2 ...+tk) =1-1μ1t1+1-1μ2t2+1-1μktk=(t1+t2 ...+tk) -t1μ1+t2μ2+tkμk=-1μi=1kti=-i=1ktiμiμ=i=1kti-i=1ktiμi
Hence, the required equivalent refractive index is i=1ktii=1ktiμi.

flag
Suggest Corrections
thumbs-up
0
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