Question

In a single slit diffraction experiment the first minimum for red light of wavelength $6600{{A}^{0}}$ coincides with the first maximum for other light of wavelength $\lambda$. The value of $\lambda$ is:

& A.2200{{A}^{0}} \\\ & B.3300{{A}^{0}} \\\ & C.4400{{A}^{0}} \\\ & D.5000{{A}^{0}} \\\ \end{aligned}$$

Answer 1

Hint: Whenever there is a coinciding condition of maxima and minima, the path difference must be the same. This will help us to solve the question.

Formula used:
$a\sin \theta =n\lambda$

Where:
a = size of a single slit
$\theta =$ angular path difference
n = integer to give the position of minima
$\lambda =$wavelength of incident light

Complete step by step answer:
Diffraction: The phenomenon of bending of light around corners of an obstacle or aperture in the path of light is called diffraction.

For the condition of minima

$a\sin \theta =n\lambda$

Where:

$n=\pm 1,\pm 2,\pm 3...$

First minima of red light is at n=1

$a\sin \theta ={{\lambda }_{red}}$

$\sin \theta =\dfrac{{{\lambda }_{red}}}{a}$

Maxima of unknown wavelength lies between the first and second minima.

First minima:

$a\sin \theta =\lambda$ …….(1)

Second minima

$a\sin \theta =2\lambda$ …….(2)

Therefore first maxima will lie in between two consecutive minima;

Adding (1) and (2)
$2a\sin \theta =3\lambda$
$a\sin \theta =\dfrac{3}{2}\lambda$
$\sin \theta =\dfrac{3\lambda }{2a}$

And we know that $\sin \theta =\dfrac{{{\lambda }_{red}}}{a}$, ( angle is same because two wavelengths are coinciding)

& \dfrac{{{\lambda }_{red}}}{a}=\dfrac{3\lambda }{2a} \\\ & \lambda =\dfrac{2{{\lambda }_{red}}}{3} \\\ & \lambda =\dfrac{2\times 6600{{A}^{0}}}{3} \\\ & \\\ \end{aligned}$$ Answer is option (C). $$4400{{A}^{0}}$$ Note: There is an alternate option to solve this question: In diffraction, Path difference for minima is given by $$\Delta x=\dfrac{n\lambda a}{D}$$ Where D is the distance between slit and screen. Therefor path difference of red light for first minima (n=1) is $$\Delta x=\dfrac{{{\lambda }_{red}}a}{D}$$ And, Path difference for maxima is given by $$\Delta x=\left( n+\dfrac{1}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D}$$ Therefore, path difference of unknown light for first maxima (n=1) $$\begin{aligned} & \Delta x=\left( 1+\dfrac{1}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D} \\\ & \Delta x=\left( \dfrac{3}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D} \\\ \end{aligned}$$ And for coinciding conditions the path difference of two lights must be the same. $$\begin{aligned} & \dfrac{{{\lambda }_{red}}a}{D}=\dfrac{3\lambda a}{2D} \\\ & \lambda =\dfrac{2{{\lambda }_{red}}}{3} \\\ & \lambda =\dfrac{2\times 6600{{A}^{0}}}{3} \\\ & \\\ \end{aligned}$$