Problems on Diffraction – Class 12 Physics

The bending of light at the edges of an obstacle whose size is comparable to the wavelength of light is called diffraction. To put it another way, it is the spreading of waves when they go through or around a barrier. Diffraction of light, as it is used to describe light, occurs more explicitly when a light wave passes by a corner or via an opening or slit that is physically smaller than the wavelength of that light, if not even smaller. The ratio of the wavelength of the light to the opening size determines how much bending occurs. The bending will essentially be undetectable if the aperture is substantially greater than the light’s wavelength. However, if the two are of similar size or are equal in size, there is a noticeable degree of bending that can be observed with the unaided eye.

Diffraction Due to Single Slit 

Diffraction is supposed to be due to interference of secondary wavelets from the exposed portion of the wavefront from the slit. Whereas in interference, all bright fringes have the same intensity. In diffraction, bright bands are of decreasing intensity. 

Condition for minimum intensity is given by,

a sin θ = nλ  (n = 1, 2, 3, …..)

where,

• a is the width of the slit
• θ is the angle of diffraction

Condition for maximum intensity is given by, 

a sinθ = (2n + 1) λ/2  (n = 1, 2, 3, ….)

where,

• a is the width of the slit
• θ is the angle of diffraction.

Intensity of Single Slit Diffraction Patterns 

The intensity decreases as we go to successive maxima away from the centre, on either side. The width of the central maxima is twice that of the secondary maxima. If this experiment is performed in liquid other than air, the width of the diffraction maxima will decrease and becomes 1/μ times. With white light, the central maximum is white and the rest of the diffraction bands are coloured.

I = I₀ [(sinβ/2) / (β/2)]² 

Here, β/2= π a sin θ /λ. So, the formula can be rewritten as,

I = I₀ [sin(π a sinθ/λ) / (πasinθ)/λ]²

where,

• θ is the angle
• I₀ is the intensity at θ = 0 (the central maximum)

The value of the first minima is calculated by using the expression of wavelength.

It is known that, asinθ = λ 

As sinθ ≈ tanθ, the value becomes,

a (y/D) = λ

y = λD/a

The width of the central maxima is given as w=2y.   

w = 2λD/a

where,

• λ is the wavelength
• D is the diameter
• a is the slit width

If lens is placed close to the slit, then D = f and w = 2fλ/a. Here f is the focal length of the lens.

Interference and Diffraction Bands 

If N interference bands are contained by the width of the central bright, then the width of the band becomes equal to Nβ or N(Dλ/d). Therefore the width of the slit is expressed as,

a = 2d/N

where,

• a is the width of the slit
• d is the diffraction value
• N is the number of bands

Difference between Interference and Diffraction

Solved Examples on Diffraction

Example 1. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 cm away. it is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. find the width of the slit.

Solution:

We know,

θ = Y/D  

θ = 2.5 ×10^-3 /1 radians

Now, asinθ = nλ

Since θ is very small, therefore sinθ =θ or, 

a = nλ/θ 

= 1×500×10^-9 / 2.5 × 10^-3 m 

= 2×10^-4 m 

= 0.2 mm

Example 2. A screen is placed 50 cm from a single slit, which is illuminated with 6000 Å light, If the distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit?

Answer: 

In case of diffraction at single slit, the position of minima is given by asinθ=nλ . Where d is the aperture size and for small values of θ.

We know, 

sinθ = y/D

a(y/D) = nλ, i.e., y = D/a(nλ)

So that, y₃ – y₁ = D/a (3λ – λ)

= D/a (2λ)

So, width of slit is,

a = 0.5×(2×6×10^-7) / 3×10^-3 

= 2×10^-4 m  

= 0.2 mm 

Example 3. In a single slit diffraction experiment, the first minimum for λ₁ = 660 nm coincides with the first maxima for wavelength λ₂ Calculate λ₂.

Answer:

Position of minima in diffraction pattern is given by; asinθ = nλ.

For first minima of λ₁ , we have

asinθ₁ = (1) λ₁ 

or, sinθ₁ = λ₁/a       ……(i) 

The first maxima approximately lies between first and second minima. For wavelength lambda its position will be,

a sinθ₂ = 3/2 (λ₂) 

Therefore, sinθ₂ = 3λ₂ / 2a ……..(ii)

The two will coincide if,

=> θ₁ = θ₂ 

=> sinθ₁ = sinθ₂  

So, λ₁ /a = 3λ₂ /2a    

=> λ₂ =2/3λ₁ 

= 2/3 × 600  

= 400 nm  

Example 4. Two slits are made one millimetre apart and the screen is placed one meter away. What should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern?

Answer:

We have, aθ=λ or θ = λ/a

Here, a is the width of each slit.

=> 10λ/d = 2λ/a

a = d/5

= 1/5

= 0.2 mm

FAQs on Diffraction

Question 1. What is the significance of diffraction?

Solution:

The bending of light gently as it travels around an object’s or opening’s edge is the basis for diffraction. It finds its applications in sound waves, telescope and microscopes.

Question 2. Explain the meaning of the single slit diffraction pattern. 

Solution:

The single-slit diffraction pattern causes the interference of light from a coherent source with itself. It allows us to examine the phenomena of light bending, or diffraction, which enables coherent light from a source to interfere with itself and generate the diffraction pattern, a recognisable pattern on the screen. Additionally, diffraction is visible when the sources are so small that they are comparable to the wavelength of light.

Question 3. What is meant by the Fringe width? Write its formula.

Solution:

The gap between two successive minima or maxima bright patches is known as the fringe width. It is denoted by the symbol β. Its formula is given as,

β = λD/d 

Question 4. Briefly write the prerequisites for constructive and destructive interference.

Solution:

For constructive interference, the path difference must be equal to an integral multiple of the wavelength in order. While for destructive interference, the path difference needs to be an odd integral multiple of a half wavelength.

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