Question

Visible light of wavelength 6000 × 10⁻⁸ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minima are at 60^o from the central maxima. If the first minimum is produced at θ₁, then θ₁ is close to

• A. ${20}^o$
• B. ${45}^o$
• C. ${30}^o$
• D. ${25}^o$

Answer 1

The correct option is D

${25}^o$

Step 1: Given data

The wavelength of light is $\lambda =6000\times {10}^{-8}cm$.

And the angle between the central maxima and the second diffraction minima will be, $\theta ={60}^o$

Step 2: Formula used

$n\lambda =d\sin \theta [whered=slitwidth,\lambda =wavelengthofthelight,\theta =angleofddiffractionminimum,n=orderoftheminima]$

Step 3: Calculating ${\theta }_1$

For a diffraction minima to occur, the product of the order of the patterns formation and the wavelength of the light will be equal to the product of slit width and the sine of the angle at which the first minimum is produced.

The second diffraction minima have been produced. Hence we can write that, $n=2$

Substituting all the given values in the equation

$$\begin{gathered} 2\lambda =d\sin {60}^o \\ d=\frac{2\lambda }{\sin {60}^o} \\ d=\frac{4\lambda }{\sqrt{3}} \end{gathered}$$

Therefore the slit width has been obtained.
Now in the second condition,
The first minima have been produced, $n=1$

Substituting the values accordingly in the equation will give,
$$\begin{gathered} 1\lambda =d\sin {\theta }_1 \\ \sin {\theta }_1=\frac{\lambda }{d} \end{gathered}$$
Now let us substitute the values of slit width in it.
$\sin {\theta }_1=\frac{\lambda }{{\displaystyle \frac{4\lambda }{\sqrt{3}}}}=\frac{\sqrt{3}}{4}$

From this, the angle will be,

${\theta }_1={\sin }^{-1}\frac{\sqrt{3}}{4}\approx {25}^o$

Hence, option D is the correct answer.