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 $$2000$$ A° 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:

Question

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  $$2000$$ A° 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:

Infinity Learn · Accepted Answer

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. $$y=0.5$$λDd 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.

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