Sunlight is used in a double-slit interference experiment. The fourth-order maxi

Question

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 490 nm occurs at an angle of d = 90degree. Thus, it is on the verge of being eliminated from the pattern because d cannot exceed 90degree in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?

Explanation / Answer

For nth order maxima (d sintheta = n lambda ) where d is the slit separation.

For 4th order maxima ( theta =90deg ) ( therefore d sin 90= 4 lambda = 4 times 490 : nm = 1960 : nm )
( =>d = 1960 : nm )
(a) For 3rd order maxima, 1960 sin theta = 3 lambda

therefore sin theta = 3 lambda/1960
For lambda > 653.3 nm , no value of theta is possible

(b) The separation should be decreased.

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