Chapter 10: Wave Optics
Key Formulas:
• Double slit: d sinθ = nλ (maxima)
• d sinθ = (n + ½)λ (minima)
• Fringe width: β = λD/d
• Single slit minima: a sinθ = nλ
Q1. In a Young's double slit experiment, the slit separation is 0.5 mm and the screen is 1 m away. If the wavelength of light used is 600 nm, find the fringe width.
Solution
β = λD/d
= (600 × 10−9)(1)/(0.5 × 10−3)
= 6 × 10−4/5 × 10−4
= 1.2 mm
Q2. In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, find the new fringe width.
Solution
β' = β/n = 0.4/(4/3) = 0.4 × 3/4
= 0.3 mm
Q3. What is the condition for destructive interference of light waves?
Solution
Path difference Δx = (2n + 1)λ/2
or Phase difference φ = (2n + 1)π
Q4. A single slit of width 0.2 mm is illuminated by light of wavelength 500 nm. Find the angular position of the first minimum.
Solution
a sinθ = nλ
For n = 1: sinθ = λ/a
= (500 × 10−9)/(0.2 × 10−3)
= 2.5 × 10−3
θ ≈ 2.5 × 10−3 rad = 0.14°
Q5. State Huygens' principle. How does it explain the rectilinear propagation of light?
Solution
Huygens' Principle: Every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the envelope of these secondary wavelets.
For rectilinear propagation: Secondary wavelets from a point source spread in all directions, but the tangent to these wavelets at any instant forms a spherical wavefront, explaining propagation in straight lines in a homogeneous medium.
Q6. What is a diffraction grating? Show that the condition for maxima is d sinθ = nλ.
Solution
A diffraction grating has a large number of equally spaced parallel slits.
For constructive interference, path difference between light from adjacent slits must be an integral multiple of λ:
d sinθ = nλ (n = 0, 1, 2, ...)
where d is the grating element (slit separation).
Higher order maxima (n > 0) are brighter than in Young's experiment because more slits contribute.
Q7. In Young's double slit experiment, the two slits are separated by 0.5 mm and the screen is 1 m away. The 5th bright fringe is at a distance of 5 mm from the central fringe. Find the wavelength of light used.
Solution
For nth bright fringe: yn = nλD/d
5 × 10−3 = 5 × λ × 1/(0.5 × 10−3)
λ = (5 × 10−3 × 0.5 × 10−3)/5
= 5 × 10−7 m = 500 nm
Q8. Prove that in Young's double slit experiment, the width of the central maximum is twice the width of other bright fringes.
Solution
Fringe width: β = λD/d
For bright fringes: yn = nλD/d
Position of 1st bright fringe: y₁ = λD/d = β
Width of central maximum: Distance between the first minima on either side.
First minima: a sinθ = λ/2 (for each slit, but in YDSE context, the central bright fringe extends from n = −1 to n = +1 minima)
Width of central maximum = y₁ − y−1 = 2β = 2 × width of other fringes
Q9. What is polarization of light? Explain how a Polaroid works.
Solution
Polarization: The phenomenon of restricting the vibrations of a transverse wave to a single plane perpendicular to the direction of propagation.
Polaroid: Uses long-chain molecules aligned in a particular direction. Light vibrating parallel to the transmission axis passes through; perpendicular components are absorbed.
When unpolarized light of intensity I₀ passes through a Polaroid, the transmitted intensity is I = I₀/2 (Malus's law for random polarization).
Q10. In a single slit diffraction experiment, the slit width is 0.2 mm and the wavelength of light is 600 nm. Find the angular position of the first three minima.
Solution
a sinθ = nλ
n = 1: sinθ₁ = λ/a = (600 × 10−9)/(0.2 × 10−3) = 3 × 10−3
θ₁ ≈ 3 × 10−3 rad = 0.17°
n = 2: θ₂ ≈ 6 × 10−3 rad = 0.34°
n = 3: θ₃ ≈ 9 × 10−3 rad = 0.52°
Q11. Unpolarized light of intensity I₀ is passed through two Polaroids whose transmission axes make an angle of 60° with each other. Find the intensity of transmitted light.
Solution
After first Polaroid: I₁ = I₀/2
By Malus's law:
I = I₁ cos²θ = (I₀/2) cos²60°
= (I₀/2)(1/2)² = (I₀/2)(1/4)
= I₀/8
Q12. How does Huygens' principle explain the laws of reflection using wave theory?
Solution
Using Huygens' construction:
1. A plane wavefront AB falls on reflecting surface XY.
2. Secondary wavelets from A reach the surface first. By the time wavelet from B reaches the surface, the one from A has traveled a distance equal to the width of the wavefront.
3. Tangent to these wavelets gives the new wavefront A'B'.
From the geometry of the construction:
Angle of incidence = Angle of reflection
This proves the first law of reflection.
Q13. In Young's double slit experiment, the fringe width is measured as 1.2 mm. If the whole apparatus is immersed in a liquid of refractive index 1.33, what will be the new fringe width?
Solution
β' = β/n = 1.2/1.33
= 1.2/(4/3) = 1.2 × 3/4
= 0.9 mm
Q14. What is Rayleigh's criterion of resolution? Define the resolving power of a telescope.
Solution
Rayleigh's criterion: Two point sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other.
Minimum angular separation: θmin = 1.22λ/d
Resolving power of telescope:
RP = 1/θmin = d/(1.22λ)
where d = aperture diameter of the objective. Higher aperture → better resolution.