Solve #2 2. If a material with the index of refraction n and the width L is plac

Question

Solve #2

2. If a material with the index of refraction n and the width L is placed in front of one slit in the double-slit experiment performed in vacuum, how do the interference conditions given by (Eq. 11) and (Eq. 12) change?

4 of 16 hys IC periment Report 2nd Semester, Year 2015 uble Slit minimum light intensity appear on the screern. Maxima of the interference pattern AL=' =mA= D (2m )(m: integer) 2 (Eq.11) slit as shown Minima of the interference pattern AL=(m AL=D= (m+ )= 2(2m + 1 ) (rn:.nteger) )--(2m +1 ) (m integer) Eq12 (Eq 12) Let us find the separation of the interference pattern, that is, the distance between adjacent maxima or minima on the screen. Since the difference of integers corresponding to adjacent maxima or minima is m=1, the separation of the interference pattern is given by ,-- (Eq 13) . By using this result, the light wavelength can be found from the separation y of the interference pattern. (2) Diffraction due to a single slit roperty. When Let us consider the diffraction phenomena due to a single slit as shown in 1, a single slit Fig. 6. e a laser as a means that a t causing the mplification by coherent laser 5, and S to een is located f two lights is y made for a

Explanation / Answer

lambda = lambda/n

this is the change in eq 11 and 12

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