Questions: Wave Optics

Q1. What is a wavefront? Answer (English): A wavefront is the surface joining all points of a wave that have the same phase at a given instant. For a point source it is a sphere centred on the source, and far from the source it can be treated as a plane. Rays are drawn perpendicular to wavefronts and show the direction in which the wave travels. Hinglish: Wavefront un sab points ka surface hai jinka phase same ho. Point source se spherical aur door se plane wavefront banti hai. Ray wavefront ke perpendicular hoti hai.

Q2. State Huygens principle. Answer (English): Every point on a wavefront behaves as a secondary source that emits spherical wavelets traveling with the wave speed. After a short time the new position of the wavefront is the common tangent surface, the envelope, of all these wavelets. The construction explains reflection and refraction of waves. Hinglish: Huygens principle: wavefront ka har point secondary wavelets bhejta hai, aur ek time baad unka envelope naya wavefront deta hai. Isse reflection aur refraction samajh aate hain.

Q3. What are coherent sources? Answer (English): Two sources are coherent when they emit waves of the same frequency with a constant phase difference. The interference pattern then stays steady in time. Independent lamps are not coherent because their emissions have random phases, so any pattern averages out. Hinglish: Coherent sources same frequency aur constant phase difference wale hote hain. Do independent lamps coherent nahi hoti kyunki unki phase random hoti hai.

Q4. Contrast the interference pattern of a double slit with the diffraction pattern of a single slit, and state what controls the fringe spacing. Answer (English): Interference fringes have a common width and the intensity falls off slowly away from the centre. In a single slit diffraction pattern the central band is twice as wide as the side bands and much brighter. The fringe spacing is set by the wavelength and by the ratio of the screen distance to the slit separation. Hinglish: Double slit interference me fringes barabar width ki hoti hain aur intensity dheere kam hoti hai. Single slit diffraction me central band dugni chaudi hoti hai. Spacing wavelength aur screen distance/slit separation se tay hoti hai.

Q5. What happens to the fringe width when the whole apparatus is immersed in a liquid of refractive index n? Answer (English): The frequency stays the same but the wavelength in the liquid becomes lambda divided by n. Since the fringe width is proportional to the wavelength, the new fringe width is beta divided by n. The pattern shrinks and the fringes become narrower. Hinglish: Liquid me wavelength lambda/n ho jaata hai, isliye fringe width bhi n se divide ho jaati hai.

Q6. Distinguish between interference and diffraction in two points. Answer (English): Interference comes from two separate coherent sources and produces fringes of equal width, with the intensity falling off slowly. Diffraction comes from the many wavelets of a single wavefront at one slit and produces a central maximum much wider and brighter than the secondary maxima, with intensities falling off rapidly. Hinglish: Interference do sources se banta hai aur fringes barabar width ki hoti hain. Diffraction ek slit ke wavelets se banta hai, jisme central maximum sabse chaudi aur bright hoti hai.

Q7. Write the condition for the first secondary maximum of a single slit diffraction pattern. Answer (English): The secondary maxima lie between the minima. The first minimum is at a sine theta equal to lambda, so the first secondary maximum is near a sine theta equal to three lambda by two. That gives theta equal to the inverse sine of three lambda divided by twice the slit width. Hinglish: First minimum a sin theta = lambda par hai, isliye first secondary maximum lagbhag a sin theta = 3 lambda / 2 par hota hai.

Q8. State Malus's law. Answer (English): When polarised light of intensity I0 passes through an analyser whose axis makes an angle theta with the axis of the polariser, the transmitted intensity is I0 cosine squared theta. The intensity is maximum when the axes are parallel and zero when they are perpendicular. Hinglish: Malus law: I = I0 cos^2 theta. Axes parallel par maximum, perpendicular par zero.

Q9. Why is a polariser able to control the intensity of light? Answer (English): Ordinary light has electric field oscillations in all directions perpendicular to the beam. A polariser transmits only the component along its own axis, so the emerging beam is polarised. Rotating a second polariser changes the component it receives, which changes the transmitted intensity. Hinglish: Ordinary light me oscillations har direction me hoti hain. Polariser sirf apni axis wali component pass karta hai, isliye intensity control hoti hai.

Q10. How does the width of the central maximum change when the slit width is reduced? Answer (English): The condition for the first minimum is a sine theta equal to lambda, so a smaller slit width gives a larger angle for the first minimum. The central maximum therefore becomes wider as the slit is made narrower, and the diffraction pattern spreads out more. Hinglish: Slit chhoti karne par first minimum ka angle bada ho jaata hai, isliye central maximum chaudi ho jaati hai.

Marks pattern for this chapter

One-mark items test a definition, the fringe width formula or a qualitative change such as what happens in a liquid. Two and three mark items ask for a path difference, an intensity expression or a single slit numerical. Five-mark items combine differences between interference and diffraction with a numerical on the double slit. Writing the formula, the substitution and the final unit covers each type.

Common mistakes in this chapter

Writing the wavelength in nanometres inside the fringe formula without converting to metres is the most frequent error. Another is to use the vacuum wavelength after the apparatus has been placed in a liquid. In the single slit formula, the slit width and the wavelength must be in the same unit. In polarisation, forgetting that the first polariser already halves the intensity of unpolarised light gives an answer that is too large by a factor of two.

A final check for every numerical: confirm that the angle comes out small enough for the sine and the angle in radians to be interchangeable, and that the final unit matches the quantity asked for.

Quiz below: five multiple choice questions on wavefronts, coherence, fringe width, diffraction and Malus's law. Attempt them after revising the formula sheet and the three topic pages. Include the unit in every final answer.