Question

Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
OR
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?

Answer 1

1) Mirror equation gives $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$⟹\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

It is given that $2f<u<f$(since both $u$ and $f$ are negative)

$⟹\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

$⟹-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$

Adding $\frac{1}{f}$ on both sides gives,

$\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<\frac{1}{f}-\frac{1}{f}$

$⟹\frac{1}{2f}<\frac{1}{v}<0$

$⟹v<0$, the image lies beyond 2f. The image is real because $v$ is negative.

2) Let the angle the first polaroid makes with the direction of vibrations of electric vector in the incident light be $\theta$.

Thus the intensity after passing through the first polaroid, according to Malus' Law is $I_1=I_{0}co{s}^2\theta$

Now the next polaroid is perpendicular to the first polaroid. Hence the angle that it forms with the electric vector's vibrations is ${90}^{\circ }-\theta$.

Thus the intensity after passing through the second polaroid=$I_2=I_{1}co{s}^2({90}^{\circ }-\theta )$

$=I_{1}si{n}^2\theta =I_{0}si{n}^2\theta co{s}^2\theta$

This value is maximum for $\theta ={45}^{\circ }$