Chapter 24, Problem 57P

(a)

To determine

Angle at which the intensity of the light will be reduced to the given value.

Answer to Problem 57P

  $35.3°$ .

Explanation of Solution

Given:

The ratios of initial and final intensity, $\frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{\text{1}}{\text{3}}$

Formula used:

It is known that: $\frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{{\text{cos}}^{\text{2}}\text{θ}}{2}$

Where,

  $I_0$ is initial intensity.

  ${\text{I}}_2$ is intensity after passing the second axes of Polaroid.

  $\theta$ is an angle.

Calculation:

  $I_0$ is initial intensity and ${\text{I}}_2$ will be second intensity after passing the polaroids, then angle $\theta$ will be.

  $\because \frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{\text{1}}{\text{3}}$

  $\therefore \frac{1}{3}=\frac{{\text{cos}}^{\text{2}}\text{θ}}{2}$

  $\therefore {\cos }^2\text{θ}=\frac{2}{3}$

  $\therefore \text{θ}=35.3°$

(b)

To determine

The angle at which the intensity of the light will be reduced to given value.

Answer to Problem 57P

  $63.4°$ .

Explanation of Solution

Given:

The ratios of initial and final intensity, $\frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{\text{1}}{\text{10}}$

Formula used:

It is known that: $\frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{{\text{cos}}^{\text{2}}\text{θ}}{2}$

Where,

  $I_0$ is initial intensity.

  ${\text{I}}_2$ is intensity after passing the second axes of Polaroid.

  $\theta$ is an angle.

Calculation:

  $I_0$ is initial intensity and ${\text{I}}_2$ will be second intensity after passing the Polaroids, then angle $\theta$ will be.

  $\because \frac{{\text{I}}_{\text{2}}}{{\text{I}}_{\text{0}}}=\frac{\text{1}}{\text{10}}$

  $\therefore \frac{\text{1}}{\text{10}}=\frac{{\text{cos}}^{\text{2}}\text{θ}}{2}$

  $\therefore {\cos }^2\text{θ}=\frac{2}{\text{10}}=\frac{1}{5}$

  $\therefore \text{θ}=63.4°$