Question 246: Waves Optics

Question

A single slit diffraction pattern is formed with light of wavelength $6384 \mathop {\rm{A}}\limits^{\rm{o}}$. 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

Step 1. Condition for secondary maxima
For a single slit of width $a$, the diffraction pattern has intensity minima at:
$ a \sin	heta = m\lambda,  \quad m= \pm1, \pm2, \pm3, \ldots $
Between two minima, there is a secondary maximum. The approximate positions of secondary maxima are given by:
$ a \sin	heta \approx (m + 	frac{1}{2})\lambda, \quad m=1,2,3, \ldots $
Thus:
1st secondary max: $a\sin	heta \approx 1.5 \lambda$
2nd secondary max: $a\sin	heta \approx 2.5 \lambda$
3rd secondary max: $a\sin	heta \approx 3.5 \lambda$
etc.
Step 2. Apply the given condition
For wavelength $\lambda_1 = 6384 \,	ext{ÃÂ}$:
2nd secondary maximum Ã¢ÂÂ $a \sin	heta = 2.5 \lambda_1$.
For wavelength $\lambda_0$:
3rd secondary maximum Ã¢ÂÂ $a \sin	heta = 3.5 \lambda_0$.
These coincide at the same angular position $	heta$, so:
$ 2.5 \lambda_1 = 3.5 \lambda_0 $
Step 3. Solve for $\lambda_0$
$ \lambda_0 = \frac{2.5}{3.5} \lambda_1 = \frac{5}{7}\lambda_1 $
$ \lambda_0 = \frac{5}{7}	imes 6384 \approx 4560 \,	ext{ÃÂ} $
Ã¢ÂÂ Final Answer:
$ \lambda_0 = 4560 \,	ext{ÃÂ} $
Correct option: B.

Answer

$4242 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$4560 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$5474 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$6384 \mathop {\rm{A}}\limits^{\rm{o}}$

Waves Optics

A single slit diffraction pattern is formed with light of wavelength 6384 \mathop {\rm{A}}\limits^{\rm{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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