Chapter 24, Problem 87GP

(a)

To determine

The angle between the emerging beams.

Answer to Problem 87GP

  $4.8°$

Explanation of Solution

Given data:

Refractive index of air, ${\text{n}}_{\text{air}}=1.00$

Refractive index of prism, ${\text{n}}_{\text{prism}}=1.652$

Formula used:

  ${\text{n}}_{\text{air}}\sin {\theta }_{\text{a}}={\text{n}}_{\text{prism}}\sin {\theta }_{\text{b}}$

Where,

  ${\text{n}}_{\text{air}}$ is refractive index of air.

  ${\text{n}}_{\text{prism}}$ is refractive index of prism.

  $\theta$ is an angle.

Calculation:

  

For the first refraction at the first surface, ${\text{n}}_{\text{air}}\sin {\theta }_{\text{a}}={\text{n}}_{\text{prism}}\sin {\theta }_{\text{b}}$ ,

  $\left(1.00\right)\sin 45°=\left(1.652\right)\sin {\theta }_{{\text{b}}_1}$

  $\therefore {\theta }_{{\text{b}}_1}=25.34°$

Similarly,

  $\left(1.00\right)\sin 45°=\left(1.652\right)\sin {\theta }_{{\text{b}}_2}$

  $\therefore {\theta }_{{\text{b}}_2}=25.90°$

Now, the angle of incidence at the second surface, $\left(90°-{\theta }_b\right)+\left(90°-{\theta }_c\right)+\text{A}=180°$

  ${\theta }_c{}_{{}_1}=A-{\theta }_b{}_{{}_1}=60°-25.34°=34.66°$

  ${\theta }_c{}_{{}_2}=A-{\theta }_b{}_{{}_2}=60°-25.90°=34.10°$

For the refraction at the second surface: $\text{n}\sin {\theta }_c={\text{n}}_{\text{air}}\sin {\theta }_{\text{d}}$

  $\left(1.652\right)\sin 34.66°=\left(1.00\right)\sin {\theta }_{{\text{d}}_1}$

  $\therefore {\theta }_{{\text{d}}_1}=69.96°$

Similarly,

  $\left(1.619\right)\sin 34.10°=\left(1.00\right)\sin {\theta }_{{\text{d}}_2}$

  $\therefore {\theta }_{{\text{d}}_2}=65.20°$

Therefore, the angle between the emerging beams is: ${\theta }_{{\text{d}}_1}-{\theta }_{{\text{d}}_2}$

  $\therefore {\theta }_{{\text{d}}_1}-{\theta }_{{\text{d}}_2}=69.96°-65.20°=4.8°$

Conclusion: The angle between the emerging beams is $4.8°$ .

(b)

To determine

The angle between two beams when diffraction grating lines is given.

Answer to Problem 87GP

  $8.7°$

Explanation of Solution

Given data:

Number of lines, $\frac{1}{\text{d}}=6200\text{lines/cm}$

Wavelength, ${\lambda }_1=420\times {10}^{-9}\text{m}$

Wavelength, ${\lambda }_2=650\times {10}^{-9}\text{m}$

Formula used:

It is known that: $d\sin \theta =m\lambda$

Where,

  $\lambda$ is wavelength.

  $d$ is the width of slit.

  $\theta$ is an angle.

Calculation:

The angle for the first-order from, $d\sin \theta =m\lambda$ ,

  $\left\{\frac{1}{\text{d}}=6200\text{lines/cm}\right\}\left({10}^{-2}\text{m/cm}\right)\sin {\theta }_1=420\times {10}^{-9}\text{m}$

  $\therefore \sin {\theta }_1=0.2604$

  $\therefore {\theta }_1=15.09°$

Similarly,

  $\left\{\frac{1}{\text{d}}=6200\text{lines/cm}\right\}\left({10}^{-2}\text{m/cm}\right)\sin {\theta }_2=650\times {10}^{-9}\text{m}$

  $\therefore \sin {\theta }_2=0.4030$

  $\therefore {\theta }_2=23.77°$

Therefore, the angle between the first order maxima: ${\text{θ}}_{\text{2}}-{\text{θ}}_{\text{1}}$

  $\therefore {\text{θ}}_{\text{2}}-{\text{θ}}_{\text{1}}=\left(23.77-15.09\right)°=8.7°$

Conclusion: The required angle is $8.7°$ .