The length of plane mirror is $$1cm$$. The wavelength of source is $$5000$$ $$A0$$ . The number of maxima formed on the large screen are (P$$ is central point on $$screen)

Question

The length of plane mirror is $$1cm$$. The wavelength of source is $$5000$$ $$A0$$ . The number of maxima formed on the large screen are (P$$ is central point on $$screen)

Infinity Learn · Accepted Answer

An image S' will form at $$5mm$$ behind the mirror. Hence, Path difference at P $$=$$ ∆x $$=$$ S'P $$-$$ SP $$=$$ $$10$$ mm  Also,  ∆x $$=$$ nλ   [Condition for minima, as mirror introduces phase difference of π] ⇒$$n=$$ ∆xλ$$=10$$×$$10-35$$×$$10-7$$ $$=$$ $$20$$,$$000th$$ minima  At infinity,  ∆x $$=$$ $$0$$  means minima  Hence, between lower Infinity and point P , $$19$$,$$999$$ maxima exists.  No interference occur above point P.

The length of plane mirror is 1cm. The wavelength of source is $5000 A^{0}$ . The number of maxima formed on the large screen are (P is central point on screen)

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The length of plane mirror is 1cm. The wavelength of source is 5000 A0 . The number of maxima formed on the large screen are (P is central point on screen)

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The length of plane mirror is 1cm. The wavelength of source is $5000 A^{0}$ . The number of maxima formed on the large screen are (P is central point on screen)