## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 33 - Wave Optics - Exercises and Problems - Page 955: 20

#### Answer

(a) $\frac{d}{\lambda} = 2$ (b) $\frac{d}{\lambda} = 1.15$ (c) $\frac{d}{\lambda} = 1$

#### Work Step by Step

We can write the general condition for a minimum: $d~sin~\theta_m = m~\lambda$ $\frac{d}{\lambda} = \frac{m}{sin~\theta_m}$ (a) We can find the slit-to-wavelength ratio when $\theta_1 = 30^{\circ}$: $\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$ $\frac{d}{\lambda} = \frac{1}{sin~30^{\circ}}$ $\frac{d}{\lambda} = 2$ (b) We can find the slit-to-wavelength ratio when $\theta_1 = 60^{\circ}$: $\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$ $\frac{d}{\lambda} = \frac{1}{sin~60^{\circ}}$ $\frac{d}{\lambda} = 1.15$ (c) We can find the slit-to-wavelength ratio when $\theta_1 = 90^{\circ}$: $\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$ $\frac{d}{\lambda} = \frac{1}{sin~90^{\circ}}$ $\frac{d}{\lambda} = 1$

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