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Light of wavelength 589 nm in air enters water (n = 1.33). Find its wavelength, frequency and speed in water.
Given: λ₀ = 589 nm; n = 1.33
Use: λ = λ₀/n; v = c/n; ν unchanged
λ = 589/1.33 = 443 nm
ν = c/λ₀ = 3 × 10⁸ / 589 × 10⁻⁹ = 5.09 × 10¹⁴ Hz (same in water)
v = 3 × 10⁸/1.33 = 2.26 × 10⁸ m/s
Answer: 443 nm; 5.09 × 10¹⁴ Hz; 2.26 × 10⁸ m/s
Find the speed of light in glass of refractive index 1.5 and the wavelength in glass of light of frequency 5 × 10¹⁴ Hz.
Given: n = 1.5; ν = 5 × 10¹⁴ Hz
Use: v = c/n; λ = v/ν
v = 3 × 10⁸/1.5 = 2 × 10⁸ m/s
λ = 2 × 10⁸ / 5 × 10¹⁴ = 4 × 10⁻⁷ m = 400 nm
Answer: 2 × 10⁸ m/s; 400 nm
Light has speed 3 × 10⁸ m/s in air and 2.25 × 10⁸ m/s in a liquid. Find n and the angle of refraction for incidence at 53°.
Given: v₁ = 3 × 10⁸; v₂ = 2.25 × 10⁸
Use: n = v₁/v₂ = sin i / sin r
n = 1.333
sin r = sin 53° / 1.333 = 0.7986/1.333 = 0.599, r = 36.8°
Answer: n = 4/3; r ≈ 37°
In a double-slit experiment d = 0.2 mm, D = 1.0 m and λ = 500 nm. Find the fringe width.
Given: d = 0.2 mm; D = 1 m; λ = 500 nm
Use: β = λD/d
β = 500 × 10⁻⁹ × 1 / 0.2 × 10⁻³ = 2.5 × 10⁻³ m
Answer: 2.5 mm
Slits 0.28 mm apart, screen 1.4 m away; the fourth bright fringe is 1.2 cm from the central one. Find λ.
Given: d = 0.28 mm; D = 1.4 m; y₄ = 1.2 cm
Use: y_n = nλD/d
β = 1.2/4 = 0.3 cm = 3 × 10⁻³ m
λ = βd/D = 3 × 10⁻³ × 0.28 × 10⁻³ / 1.4 = 6 × 10⁻⁷ m
Answer: 600 nm
The experiment above (β = 2.5 mm in air) is performed in water (n = 4/3). Find the new fringe width.
Given: β = 2.5 mm; n = 4/3
Use: β′ = β/n
β′ = 2.5 × 3/4 = 1.875 mm
Answer: 1.9 mm (about)
With λ = 600 nm, D = 1.5 m, d = 0.3 mm, find the distance of the third bright fringe and the second dark fringe from the centre.
Given: λ = 600 nm; D = 1.5 m; d = 0.3 mm
Use: β = λD/d = 3 mm
Third bright: 3β = 9 mm
Second dark: (1 + ½)β = 1.5 × 3 = 4.5 mm
Answer: 9 mm; 4.5 mm
In a double-slit pattern what is the path difference at the fifth bright and the fifth dark fringe from the centre (λ = 500 nm)?
Given: λ = 500 nm
Use: Δ = nλ, (n − ½)λ
Fifth bright: 5λ = 2500 nm
Fifth dark: 4.5λ = 2250 nm
Answer: 2500 nm; 2250 nm
A transparent sheet of n = 1.5 and thickness 4 μm covers one slit. By how many fringes does the pattern shift for λ = 500 nm?
Given: n = 1.5; t = 4 μm; λ = 500 nm
Use: N = (n − 1)t/λ
N = 0.5 × 4 × 10⁻⁶ / 500 × 10⁻⁹ = 4
Answer: 4 fringes
Find the angular separation of adjacent fringes for d = 1 mm and λ = 500 nm.
Given: d = 1 mm; λ = 500 nm
Use: θ = λ/d
θ = 500 × 10⁻⁹ / 10⁻³ = 5 × 10⁻⁴ rad
Answer: 5 × 10⁻⁴ rad
The intensity at a point where the path difference is λ is K. What is it where the path difference is λ/3?
Given: Δ₁ = λ; Δ₂ = λ/3
Use: I = I_m cos²(πΔ/λ)
At Δ = λ: I = I_m cos²π = I_m = K
At Δ = λ/3: I = K cos²(π/3) = K/4
Answer: K/4
Two coherent sources have intensities in the ratio 9 : 1. Find I_max/I_min.
Given: I₁ : I₂ = 9 : 1
Use: (√I₁ + √I₂)² / (√I₁ − √I₂)²
(3 + 1)² / (3 − 1)² = 16/4 = 4
Answer: 4
The maximum intensity on the screen is K. Find the intensity where the path difference is λ/6.
Given: Δ = λ/6
Use: I = K cos²(πΔ/λ)
I = K cos²(π/6) = K × 3/4
Answer: 3K/4
Each of two slits alone gives intensity I₀ at the screen. Find the maximum and minimum intensities.
Given: I₁ = I₂ = I₀
Use: I_max = (√I₁ + √I₂)²
I_max = 4I₀
I_min = 0
Answer: 4I₀; 0
Light of 650 nm and 520 nm is used in a double-slit set-up with D/d = 600. Find the third bright fringe for 650 nm, and the least distance where bright fringes of both coincide.
Given: λ₁ = 650 nm; λ₂ = 520 nm; D/d = 600 (e.g. D = 1.2 m, d = 2 mm)
Use: y = nλD/d
y₃ = 3 × 650 × 10⁻⁹ × 600 = 1.17 mm
n₁ × 650 = n₂ × 520 gives n₁ : n₂ = 4 : 5
y = 4 × 650 × 10⁻⁹ × 600 = 1.56 mm
Answer: 1.17 mm; 1.56 mm
A slit 0.1 mm wide is lit by light of 500 nm. Find the angle of the first minimum.
Given: a = 0.1 mm; λ = 500 nm
Use: sin θ = λ/a
sin θ = 500 × 10⁻⁹ / 10⁻⁴ = 5 × 10⁻³
θ = 5 × 10⁻³ rad = 0.29°
Answer: 5 × 10⁻³ rad
A slit of width 0.2 mm is lit by 600 nm light, with the screen 1 m away. Find the width of the central maximum.
Given: a = 0.2 mm; λ = 600 nm; D = 1 m
Use: width = 2λD/a
2 × 600 × 10⁻⁹ × 1 / 0.2 × 10⁻³ = 6 × 10⁻³ m
Answer: 6 mm
For a slit of width 4 μm and λ = 500 nm, find sin θ for the second minimum.
Given: a = 4 μm; λ = 500 nm
Use: a sin θ = 2λ
sin θ = 2 × 0.5/4 = 0.25
Answer: 0.25
Unpolarised light of intensity I₀ passes through two polaroids with axes at 60°. Find the transmitted intensity.
Given: θ = 60°
Use: I = (I₀/2) cos²θ
I = (I₀/2)(1/4) = I₀/8
Answer: I₀/8
Plane polarised light of intensity I₀ falls on an analyser at 30° to its polarisation. Find the emergent intensity.
Given: θ = 30°
Use: I = I₀ cos²θ
I = I₀ × 3/4
Answer: 0.75 I₀
Unpolarised light I₀ passes through polaroids with axes at 0°, 45° and 90°. Find the final intensity.
Given: 0°, 45°, 90°
Use: Apply Malus' law at each step
After P₁: I₀/2
After P₂: (I₀/2) cos²45° = I₀/4
After P₃: (I₀/4) cos²45° = I₀/8
Answer: I₀/8
At what angle between the polarisation and the analyser axis is the intensity a quarter of the incident intensity?
Given: I = I₀/4
Use: cos²θ = 1/4
cos θ = 1/2
θ = 60°
Answer: 60°
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Frequency: ν = c/λ = 3 × 10⁸ / 589 × 10⁻⁹ = 5.09 × 10¹⁴ Hz. This is the same for the reflected and the refracted light.
(a) Reflected light stays in air: λ = 589 nm, ν = 5.09 × 10¹⁴ Hz, speed = 3 × 10⁸ m/s.
(b) Refracted light: ν = 5.09 × 10¹⁴ Hz; speed v = c/n = 3 × 10⁸/1.33 = 2.26 × 10⁸ m/s; wavelength λ = v/ν = 2.26 × 10⁸ / 5.09 × 10¹⁴ = 443 nm (= 589/1.33).
(a) Spherical (diverging).
(b) Plane: rays from the focus come out parallel.
(c) Plane: the star is so far away that the very small portion of its spherical wavefront which reaches the Earth is practically flat.
(a) v = c/n = 3.0 × 10⁸ / 1.5 = 2.0 × 10⁸ m/s.
(b) No. The refractive index of glass depends on the colour (it is larger for violet than for red), so the speed in glass differs for different colours. Violet light travels slower.
For the fourth bright fringe y₄ = 4λD/d. So the fringe width β = 1.2 cm / 4 = 0.3 cm = 3 × 10⁻³ m.
λ = βd/D = 3 × 10⁻³ × 0.28 × 10⁻³ / 1.4 = 6 × 10⁻⁷ m = 600 nm.
The intensity at a point with phase difference φ is I = I_m cos²(φ/2), where φ = 2πΔ/λ.
For Δ = λ: φ = 2π, so I = I_m cos²π = I_m = K (a maximum).
For Δ = λ/3: φ = 2π/3, I = K cos²(π/3) = K × (1/2)² = K/4.
Note on the data: the slit separation d and screen distance D are not given in this printed version. The answer depends on D/d. For the usual values of this exercise, d = 2 mm and D = 1.2 m (D/d = 600), the answers are as follows.
(a) y₃ = 3λD/d = 3 × 650 × 10⁻⁹ × 600 = 1.17 × 10⁻³ m = 1.17 mm. In general y₃ = 3 × (650 nm) × (D/d).
(b) Bright fringes coincide when n₁(650) = n₂(520), i.e. n₁ : n₂ = 520 : 650 = 4 : 5. So the least coincidence is at the 4th fringe of 650 nm (= 5th of 520 nm): y = 4 × 650 × 10⁻⁹ × 600 = 1.56 mm. In general y = 4 × (650 nm) × (D/d).