Explanation 1) Concept: When light waves encounter an edge or an obstacle of size comparable to the wavelength of the waves, those waves spread out and undergo interference and give diffraction pattern. And by using diffraction condition, one can find out wavelengths that pass through the slit. 2) Formula: i) Condition for diffraction maxima d s i n θ = m λ 3) Given: i) A grating has 1 d = 350 r u l i n g s / m m ii) This grating is illuminated at normal incidence by white light. iii) A spectrum is formed on a screen D = 30 cm from the grating. iv) Side length of square hole, 10.0 m m is cut on the screen. v) Inner edge of square hole from the central maximum is, 50.0 m m , and is parallel to it. 4) Calculations: a) d for given grating d = 1 350 m m = 0.00285 m m = 0.00285 × 10 6 n m By using diffraction condition, we get m = d s i n θ λ = 0.00285 × 10 6 n m × s i n θ λ We have t a n θ = 5.0 c m 30 c m = 0.1666 tan - 1 ⁡ 0

11th Edition

ISBN: 9781119306887

Author: Halliday

Publisher: WILEY

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Question

Chapter 36, Problem 53P

To determine

To find

a) the shortest wavelength of the light that passes through the hole.

b) the longest wavelength of the light that passes through the hole.

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Chapter 36, Problem 52P

Chapter 36, Problem 54P

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Chapter 36 Solutions

FUND.OF PHYSICS

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To find a) the shortest wavelength of the light that passes through the hole. b) the longest wavelength of the light that passes through the hole.

Solution Summary: The author explains that light waves encounter an edge or an obstacle of size comparable to the wavelength of the waves and undergo interference and give diftion pattern.

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To find a) the shortest wavelength of the light that passes through the hole. b) the longest wavelength of the light that passes through the hole.

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Chapter 36 Solutions

To find

a) the shortest wavelength of the light that passes through the hole.

b) the longest wavelength of the light that passes through the hole.