Beam remains Parallel!

Classical Mechanics Level 5

Two converging lenses of same aperture size are placed with their principal axis coinciding. Their focal lengths are in the ratio $K$ . When a light beam is incident parallel to their common principal axis, it is observed that final emergent beam is also parallel to their common principal axis. It is observed that when beam is incident from left intensity recorded on screen is $I_1$ , but when positions of lenses are interchanged, the same point on screen records an intensity $I_2$ . Then $\large \frac{I_1}{I_2} = \frac{1}{K^x}$ where $x$ can have 2 values. Find the sum of the two values of $x$ .

The answer is 0.00.

2 solutions

Tijmen Veltman
Jun 24, 2015

Since no information is given to specifically distinguish $I_1$ and $I_2$ , they are interchangable. Therefore if $x$ is a solution, so is $-x$ :

$\frac{I_2}{I_1}=\left(\frac{I_1}{I_2}\right)^{-1}=\left(\frac1{K^x}\right)^{-1}=\frac1{K^{-x}}$ .

Hence the sum of the two values must be $x+(-x)=\boxed{0}$ .

Aniket Sanghi
Sep 28, 2016

This q can directly be solved (in JEE) as @Tijmen Veltman said

Actual way can be (with one approximation of source rays)

Take a spherical parallel line intersecting and by symmetry you will see that they form a circle of some radius on screen which is found to be different for both situations and $I \propto \frac {1}{r^2 }$

Solving this we get x = 4 , -4

This question could have been interesting if we would have been told to find value of x with ${I}_{1}$ $and$ ${I}_{2}$ cases known . And some information about source rays! :)

i was thinking some information is required. so i did by the above way (if x is solution then -x is also)

Prakhar Bindal - 4 years, 8 months ago

Should not x be 2 and -2?

raj abhinav - 1 year ago

Beam remains Parallel!

The answer is 0.00.

2 solutions

0 pending reports