This MCQ module is based on: Ydse Diffraction Polarisation
Ydse Diffraction Polarisation
This assessment will be based on: Ydse Diffraction Polarisation
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Ydse Diffraction Polarisation
10.5 Young's Double-Slit Experiment (YDSE)
In 1801, Thomas Young did what Newton's corpuscles could not explain: he demonstrated the interference of light. A narrow source S illuminates a single slit (to create coherence), which then lights two close, parallel slits \(S_1\) and \(S_2\). These two slits act as coherent sources, and their waves overlap on a screen placed a distance \(D\) away, producing alternating bright and dark bands called fringes.
Path Difference and Fringe Width
For a point \(P\) at height \(y\) on the screen (with \(d,y\ll D\)), the path difference between the two rays is approximately:
\[\Delta x=S_2P-S_1P\approx\frac{y\,d}{D}\]Bright fringes (maxima) occur where \(\Delta x=n\lambda\):
\[y_n^\text{br}=\frac{n\lambda D}{d},\qquad n=0,\pm1,\pm2,\dots\]Dark fringes (minima) occur where \(\Delta x=(n+\tfrac12)\lambda\):
\[y_n^\text{dk}=\left(n+\tfrac12\right)\frac{\lambda D}{d}\]The spacing between consecutive bright (or dark) fringes — the fringe width — is a hallmark result:
\[\boxed{\;\beta=\frac{\lambda D}{d}\;}\]The intensity variation across the screen is
\[I=4I_0\cos^2\!\left(\frac{\pi y d}{\lambda D}\right)\]so fringes are equally spaced and have equal brightness (ideal case).
10.6 Diffraction
Diffraction is the bending of waves around obstacles or through apertures comparable in size to the wavelength. It is another direct proof of the wave nature of light.
Single-Slit Diffraction
When monochromatic light of wavelength \(\lambda\) passes through a slit of width \(a\), Huygens' wavelets from every point of the slit interfere. The screen shows a wide central maximum flanked by much weaker secondary maxima.
- Minima: \(a\sin\theta=n\lambda\), \(n=\pm1,\pm2,\dots\)
- Secondary maxima: \(a\sin\theta\approx\left(n+\tfrac12\right)\lambda\)
- Central maximum width (between first minima on either side) on a screen distance \(D\) away: \(\;W=\dfrac{2\lambda D}{a}\).
YDSE vs Single-Slit Diffraction
| Feature | YDSE (interference) | Single slit (diffraction) |
|---|---|---|
| Fringe spacing | Equal (\(\beta=\lambda D/d\)) | Central max twice as wide as side fringes |
| Intensities | All bright fringes ~equal | Central max dominates; side max much weaker |
| Condition for minima | \((n+\tfrac12)\lambda\) | \(a\sin\theta=n\lambda\) |
| Number of fringes | Many observable | Only a few side bands visible |
Resolving Power — Rayleigh's Criterion
Two point sources are just resolved by a circular aperture of diameter \(D\) when their angular separation equals
\[\Delta\theta_\text{min}=\frac{1.22\lambda}{D}\]Larger apertures (telescopes, microscopes) and shorter wavelengths give finer resolution.
10.7 Polarisation
Light is a transverse EM wave — the electric field \(\vec E\) vibrates perpendicular to the direction of propagation. Ordinary sunlight is unpolarised: \(\vec E\) rapidly and randomly changes direction in the plane perpendicular to the ray.
In polarised light, \(\vec E\) vibrates in only one fixed plane.
Malus's Law
If already polarised light of intensity \(I_0\) falls on a polariser whose axis makes an angle \(\theta\) with the light's polarisation, the transmitted intensity is:
\[\boxed{\;I=I_0\cos^2\theta\;}\]Polarisation by Reflection — Brewster's Law
When unpolarised light falls on glass at a special angle \(\theta_B\), the reflected ray is completely polarised (perpendicular to the plane of incidence). The condition is that the reflected and refracted rays are perpendicular to each other, giving:
\[\tan\theta_B=n,\qquad \theta_B+\theta_r=90°\]Applications of polarisation: polaroid sunglasses cut horizontal glare from roads and water; LCDs use polarising layers to switch pixels; 3-D cinema glasses use perpendicular polarisations per eye; photographers use polarising filters to deepen sky colour; engineers use "photo-elastic" polarisation to map stress in transparent models.
Hold one polaroid sheet in front of a lamp and rotate it — brightness is roughly constant.
Interactive — YDSE Fringe Calculator
Enter wavelength, screen distance and slit separation. The output gives the fringe width \(\beta=\lambda D/d\) and paints the resulting pattern.
Worked Examples
In YDSE, \(\lambda=600\) nm, \(D=1.2\) m, \(d=0.4\) mm. Find \(\beta\) and the position of the 3rd bright fringe.
Same setup as Ex 1. Where is the 2nd dark fringe (\(n=1\) in \((n+\tfrac12)\))?
The apparatus of Ex 1 is submerged in water of index 1.33. What is the new fringe width?
Light of 500 nm passes through a slit of width 0.1 mm; screen is 1 m away. Find the angular width of the central maximum and its linear width.
Refractive index of crown glass = 1.52. Compute Brewster's angle and the angle of refraction.
Polarised light of intensity 80 W/m² is incident on a polariser whose axis makes 30° with the E-vector. Find the transmitted intensity.
A telescope has aperture 20 cm. Minimum angular separation for 550 nm light?
Competency-Based Questions
Q1. In YDSE, if the slit separation is doubled while \(\lambda\) and \(D\) stay the same, the fringe width:
Q2. In single-slit diffraction, the first minimum occurs when:
Q3. (Short answer) Why does replacing a YDSE apparatus in water reduce the fringe width?
Q4. (Long answer) State Brewster's law and explain why the reflected light at Brewster's angle is completely polarised.
Q5. (HOT) Two polaroids are crossed (90° apart). A third is inserted between them at 45°. Find the transmitted intensity if unpolarised light of intensity \(I_0\) falls on the first.
Assertion–Reason Questions
Options: (A) Both true, R correct explanation. (B) Both true, R not correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: In YDSE, the central fringe is bright for all wavelengths.
Reason: At the central point both paths are equal, giving zero path difference for every colour.
Assertion: Sound waves can be polarised but light cannot.
Reason: Light waves are transverse while sound waves are longitudinal.
Assertion: The resolving power of a microscope increases with shorter wavelength.
Reason: The minimum resolvable angle follows \(\Delta\theta\propto\lambda/D\).
Frequently Asked Questions - Ydse Diffraction Polarisation
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🎯 Practise Physics
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Board exam sample papers
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E