Question 301: Waves Optics

Question

Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking place at a single slit of aperture 2.4 mm . The distance between the slit and screen is 2 m . The separation between the positions of first secondary maximum of the diffraction pattern obtained in the two cases is

Accepted Answer

To calculate the separation between the first secondary maxima for two different wavelengths of sodium light in a diffraction pattern, we use the formula for the position of the first secondary maximum in a single slit diffraction setup:
$ \sin 	heta = (2n + 1) \frac{\lambda}{2a} = \frac{3\lambda}{2a} $
For small angles, we can approximate $\sin 	heta \approx 	an 	heta = 	heta$. Therefore, the position $x$ of the first secondary maximum on the screen is given by:
$ x = \frac{3\lambda D}{2a} $
Here, $\lambda$ is the wavelength of light, $D$ is the distance from the slit to the screen, and $a$ is the aperture of the slit.
To find the separation $\Delta x$ between the positions of the first secondary maxima for two different wavelengths, $\lambda_1 = 590 \, 	ext{nm}$ and $\lambda_2 = 596 \, 	ext{nm}$, use:
$ \Delta x = \frac{3D(\lambda_2 - \lambda_1)}{2a} $
Substituting the given values:
$D = 2 \, 	ext{m}$
$\lambda_1 = 590 	imes 10^{-9} \, 	ext{m}$
$\lambda_2 = 596 	imes 10^{-9} \, 	ext{m}$
$a = 2.4 	imes 10^{-3} \, 	ext{m}$
we get:
$ \Delta x = \frac{3 	imes 2 	imes (596 - 590) 	imes 10^{-9}}{2 	imes 2.4 	imes 10^{-3}} $
Simplifying this:
$ \Delta x = \frac{3 	imes 2 	imes 6 	imes 10^{-9}}{4.8 	imes 10^{-3}} = 7.5 	imes 10^{-6} \, 	ext{m} $
Thus, the separation between the positions of the first secondary maximum for the two wavelengths is $7.5 	imes 10^{-6} \, 	ext{m}$.

Answer

$7.5 	imes 10^{-6} \mathrm{~m}$

Answer

$7.5 	imes 10^{-9} \mathrm{~m}$

Answer

$ 2.5 	imes 10^{-6} \mathrm{~m}$

Answer

$5.0 	imes 10^{-6} \mathrm{~m}$

Waves Optics

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