Explanation Given data: Wavelength of blue light, λ blue =500 nm Wavelength of red light, λ red =700 nm Refractive index of film n film = 1.35 Refractive index of film n water = 1.33 Formula Used: Using formula for constructive interference: - 2 D = ( 2 m − 1 ) n 1 λ 1 2 n 2 where D = t Calculation: As it has been given that white light illuminates a thin film ( n film = 1.35 ) which is normal to thesurface, and we alsoobtained that both the lights of different color, blue light ( λ blue = 500 nm )and red light λ red = 700 nm are reflected strongly. Let us consider that the filmhas a thickness t and is floating on top of water ( n water = 1.33 ). Since the film has a higher refractive index value higher than water,interference which is constructive is given by below equation: 2 D = ( 2 m − 1 ) n 1 λ 1 2 n 2 where D =t. The above obtained path difference reflects strongly two light wavelengths;thus, we will have two expressions with two different value of integers, m blue for the blue light and m red for the red light. Thus, puttingboth the thicknesses equivalent, then we can solvefor the smallest integers that satisfy the resultant relationship. Thethickness of the film which is minimum can be found by putting the value of integer again in the relationship of constructiveinterference. 2 D = ( 2 m − 1 ) n 1 λ 1 2 n 2 Blue light: 2 D = ( 2 m blue − 1 ) n air λ blue 2 n film Red light: 2 D = ( 2 m red − 1 ) n air λ red 2 n

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ISBN: 9781464196393

Author: Freedman

Publisher: MAC HIGHER

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Chapter 23, Problem 63QAP

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The minimum thickness of the film.

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Chapter 23, Problem 64QAP

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The minimum thickness of the film.

Solution Summary: The author explains that the film has a thickness of t and is floating on top of water. Since the path difference reflects strongly two light wavelengths, we will have expressions with two different values of integers.

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