*Question*

Light is incident on a diffraction grating. The wavelength lines 600.0 nm and 601.5 nm are just resolved in the second order spectrum. How many slits of the diffraction grating are illuminated?

A 20

B 40

C. 200

D. 400

**▶️Answer/Explanation**

Ans: C

Refer: https://www.iitianacademy.com/ib-unit-9-wave-phenomena-diffraction-interference-notes/

\(R= Nm =\frac{\lambda_{avg}}{\Delta \lambda}\)

In Given question

\(\lambda_1 = 600.0 nm , \lambda_2 = 601.5 \; nm\)

Hence \(\Delta \lambda = \lambda_2-\lambda_1\)

\(=601.5-600.0 = 1.5 \times 10^{-9}\)

also \(m= 2\)

\(\lambda_{avg} =\frac{\lambda_1+\lambda_2}{2}=\frac{601.5+600.0}{2} \approx 600 \times 10^{-9}\)

Putting in equation these values

\(Nm =\frac{\lambda_{avg}}{\Delta \lambda}\)

\(\therefore N= \frac{600 \times 10^{-9}}{2 \times 1.5 \times 10^{-9}}=\frac{600}{3}=200\)

*Question*

In a double-slit interference experiment, the following intensity pattern is observed for light of wavelength *λ*.

The distance between the slits is *d*. What can be deduced about the value of the ratio \(\frac{\lambda }{d}\) and the effect of single-slit diffraction in this experiment?

**▶️Answer/Explanation**

### Markscheme

B

\(d sin\theta = m \lambda \; maxima \; for \; m= \pm 1 ,\pm 2 etc.\)

for small \(\theta\)

\(d \; \theta = m \lambda\)

or

\(\frac{\lambda}{d } =\frac{\theta}{m}\)

for \(m =1\)

\(\frac{\lambda}{d } =\theta\)

Hence in given question

\(\theta = 0.01\)

We have seen that the interference maxima occur when d sinθ = mλ . On the other hand, the condition for the first diffraction minimum is a sinθ = λ . Thus, a particular interference maximum with order number m may coincide with the first diffraction minimum. The value of m may be obtained as:

\(\frac{d \; sin \theta}{a\; sin\theta}=\frac{m \lambda}{ \lambda}\)

or

\(m = \frac{d}{a}\)

Hence \(mth \) fringe will not be seen. Hence effect of single-slit diffraction in this experiment is non negligible

*Question*

A parallel beam of monochromatic light of wavelength λ passes through a slit of width *b* and forms a diffraction pattern on a screen far from the slit. The angle at which the first diffraction minimum is f\med is *θ*.

Which of the following changes in *λ* and *b*, carried out separately, will increase the value of *θ*?

**▶️Answer/Explanation**

### Markscheme

D

we have equation for diffraction minima as

\(b sin \theta = n \lambda\)

Hence \(b sin \theta = \lambda\) (first minimum)

or

\(sin \theta \approx \theta =\frac{\lambda}{b}\)

\(\therefore \theta \uparrow \; if \lambda\uparrow or \; b \downarrow\)

*Question*

A coherent beam of light of wavelength λ is incident on a double slit. The width of the slits is small compared to their separation. An interference pattern is observed on a distant screen. O is the mid point of the screen.

There is a bright fringe at O and a bright fringe at P. Between O and P there are three dark fringes.

Which of the following is the path difference between the light from the two slits arriving at P?

A. 1.5 *λ*

B. 2 *λ*

C. 3 *λ*

D. 4 *λ*

**▶️Answer/Explanation**

### Markscheme

C

Fringe width is \(\beta \)

\(\beta\) is given by

\(\beta = \frac{\lambda D}{d}\)

In given problem , \(\bar{OP}=y=\frac{\beta }{2}+\beta+\beta+\frac{\beta }{2}\)

\(=3\beta =3 \frac{\lambda D}{d}\)

Path difference is given by

\(S_2P – S_1P= \frac{yd}{D} =3 \frac{\lambda D}{d} \times\frac{d}{D} =3 \lambda\)

*Question*

A beam of monochromatic light is incident on a diffraction grating of *N *lines per unit length. The angle between the first orders is *θ*_{1}.

What is the wavelength of the light?

A. \(\frac{{\sin {\theta _1}}}{N}\)

B. *N* sin *θ*_{1}

C. *N* sin\(\left( {\frac{{{\theta _1}}}{2}} \right)\)

D. \(\frac{{\sin \left( {\frac{{{\theta _1}}}{2}} \right)}}{N}\)

**▶️Answer/Explanation**

### Markscheme

D

\(d sin\theta = m \lambda\) for \(m= 0,1,2,3\) (maxima – line)

for \(1 st\) order m =1

\(d sin\theta = \lambda\)

Now \(\theta = \frac{\theta_1}{2} \) (\(\theta_1\) is for Both side of central maxima) also \(d=\frac{1}{N}\)

Hence

\(\lambda = \frac{sin \frac{\theta_1}{2}}{N}\)