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
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.