Question 306: Wave Optics

Question

A single slit diffraction pattern is formed with light of wavelength $6195 \mathop A\limits^o$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' $\lambda_0$ '. The value of ' $\lambda_0$ ' is

Accepted Answer

In a single slit diffraction pattern, the position of secondary maxima can be determined using the angular position equation:
$$ a \sin 	heta = (m + \frac{1}{2}) \lambda $$
where:
$ a $ is the width of the slit,
$ 	heta $ is the angle of diffraction,
$ m $ is the order number of the secondary maximum,
$ \lambda $ is the wavelength of light.
For the second secondary maximum (m = 2) of the light with wavelength $ \lambda = 6195 \mathop A\limits^o $:
$$ a \sin 	heta = \left(2 + \frac{1}{2}ight) \lambda = \frac{5}{2} 	imes 6195 $$
For the third secondary maximum (m = 3) with the unknown wavelength $ \lambda_0 $:
$$ a \sin 	heta = \left(3 + \frac{1}{2}ight) \lambda_0 = \frac{7}{2} \lambda_0 $$
Since both maxima coincide, their angular positions are equal:
$$ \frac{5}{2} 	imes 6195 = \frac{7}{2} \lambda_0 $$
Solving for $ \lambda_0 $:
$ \lambda_0 = \frac{\frac{5}{2} 	imes 6195}{\frac{7}{2}} = \frac{5 	imes 6195}{7} $
$ \lambda_0 = \frac{30975}{7} = 4425 \mathop A\limits^o $
Therefore, the value of $ \lambda_0 $ is $ 4425 \mathop A\limits^o $.
Correct Option:
Option B: $ 4425 \mathop A\limits^o $

Answer

$4180 \mathop A\limits^o$

Answer

$4425 \mathop A\limits^o$

Answer

$5330 \mathop A\limits^o$

Answer

$6235 \mathop A\limits^o$

Wave Optics

A single slit diffraction pattern is formed with light of wavelength 6195 \mathop A\limits^o. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' \lambda_0 '. The value of ' \lambda_0 ' is

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