In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength $λ'$. The value of $λ'$ is: |
440 nm 550 nm 600 nm 650 nm |
440 nm |
Here, $λ_r = 660 nm ; λ' = ?$ For diffraction minima, $a \sin θ = nλ, \sin θ =\frac{nλ}{a}$ For first minima of red, $\sin θ=\frac{(1)λ_r}{a}$ For diffraction maxima $a \sin θ = (2n + 1)\frac{λ}{2}$ ∴ For first maxima of $λ'$ $a \sin θ' =\frac{3λ'}{2};\sin θ' =\frac{3λ'}{2a}$ As the two coincide, therefore, $sin θ' = sin θ$ $\frac{3λ'}{2a}=\frac{λ_r}{a}$ or $λ' = \frac{2}{3}λ_r$ or $λ' = \frac{2}{3}(660) 440nm$ |