Questions
A concave mirror of short aperture and radius of curvature 10cm is cut into two equal parts, slightly separated and placed in front of a monochromatic line source S of wavelength as shown in figure. Consider only interference in reflected light after first reflection from mirrors. Assume direct light from S is blocked from reaching the screen.
If half of reflecting surface of upper part of mirror is painted black then choose correct option:
detailed solution
Correct option is B
Due to mirrors real images S1 and S2 are formed.Now, Half Part of Upper Mirror is only covered as shown in figure.This will lead to decrease in intensity of image S1 , but no change in positions of any image. Thus , values of d and D will remain same.Position of first order minima will also remain same. y=0.5λDd Using the formulae Imax=I1+I22,Imin=I1−I22Before covering the mirror, Imin was zero. But now, Imin becomes a non zero value . Thus we can say Imin increases.On the other hand, Imax decreases.Talk to our academic expert!
Similar Questions
In an interference arrangement, similar to Young’s double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150.0 m. The intensity is measured as a function of , where is defined as shown in the figure. If I0 is maximum intensity, then for
90 is given by
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