Orange light of wavelength $$6000$$×$$10$$−$$10$$ m illuminates a single slit of width $$0.6$$×$$10$$−$$4m$$. The maximum possible number of diffraction minima produced on both sides of the central maximum is

For minima dsinθ$$=n$$λ⇒$$sin$$θ$$=n$$λd<$$1$$ ⇒n

Diffraction of light-Wave Optics

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Orange light of wavelength $6000 \times 10^{- 10}$ m illuminates a single slit of width $0.6 \times 10^{- 4} m$ . The maximum possible number of diffraction minima produced on both sides of the central maximum is

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Solution

$\begin{array}{l} For minima dsin θ = n λ \\ \Rightarrow \sin θ = \frac{n λ}{d} < 1 \\ \Rightarrow n < \frac{d}{λ} = \frac{0.6 \times 10^{- 4}}{6000 \times 10^{- 10}} = 100 \\ T h e 100^{t h} m i n i m a c o r r e s p o n d s t o i n f i n i t y . \\ S o, o n t h e s c r e e n 99 m i n i m a o n o n e s i d e i s o b t a i n e d \\ a n d 198 o n b o t h s i d e s . \end{array}$

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Diffraction of light-Wave Optics

Remember concepts with our Masterclasses.

Orange light of wavelength 6000×10−10 m illuminates a single slit of width 0.6×10−4m. The maximum possible number of diffraction minima produced on both sides of the central maximum is

For minima dsinθ=nλ⇒sinθ=nλd<1 ⇒n<dλ=0.6×10-46000×10-10=100The 100th minima corresponds to infinity.So, on the screen 99 minima on one side is obtained and 198 on both sides.

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Orange light of wavelength $6000 \times 10^{- 10}$ m illuminates a single slit of width $0.6 \times 10^{- 4} m$ . The maximum possible number of diffraction minima produced on both sides of the central maximum is