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Learn/ 12th Class Physics/ Unit 15 /Long Questions

Unit 15: Physical Optics — Long Questions

12th Class Physics · Unit 15: Physical Optics

CRQ 15.1.Light sometimes describes as rays and sometimes as waves. Construct diagrams to explain differences in both. [Diagrams expected: (i) ray diagram — straight arrows radiating from a point source, used for geometrical optics such as reflection/refraction at mirrors and lenses; (ii) wave diagram — concentric circular/spherical wavefronts (or parallel straight wavefronts for a distant source) with rays drawn perpendicular to the wavefronts, used to explain interference and diffraction.]

Ray model Light is represented as straight lines (rays) showing the direction of travel and energy flow, drawn perpendicular to the wavefront. This geometrical picture is adequate to describe phenomena such as reflection and refraction at mirrors and lenses, where light travels in straight lines between surfaces and obstacles are much larger than the wavelength. Diagram: a point source S with several straight arrows radiating outward; at a mirror/lens the rays are shown bending according to the laws of reflection/refraction.

Wave model Light is represented as wavefronts — imaginary surfaces joining all points of the same phase (circular/spherical near a point source, or plane for a distant source) — spreading outward from the source. This model is needed to explain interference and diffraction, where the overlap and superposition of waves from different points or sources produces patterns of reinforcement and cancellation that the ray model cannot predict. Diagram: a point source S with a series of concentric circles (wavefronts) spaced one wavelength apart, with short ray arrows drawn perpendicular to the wavefronts indicating the direction of energy transfer.

Key difference the ray model shows only the direction of energy travel and cannot account for bending round obstacles or bright/dark fringe patterns; the wave model, built from wavefronts and Huygens' secondary wavelets, correctly predicts interference and diffraction, which are wave phenomena.

CRQ 15.2.Suppose white light is falling on a double slit arrangement, such that one slit is covered with red filter (400 nm) and other is with blue filter (650 nm), explain what would be observed on the screen?
CRQ 15.3.What will be the effect on width and brightness of the fringes in a double slit interference experiment if we decrease the width of one slit to half keeping all other thing same?

The fringe spacing (width), Δy = λL/d, depends only on the wavelength λ, the slit separation d and the screen distance L — none of which change, so the fringe width stays the same. However, narrowing one slit lets through less light, so the amplitude of the wave from that slit becomes smaller than from the other slit. The two interfering waves then have unequal amplitudes: the bright fringes become dimmer (since perfect constructive interference of unequal amplitudes gives less total amplitude than with equal amplitudes), and the dark fringes are no longer perfectly dark (since the amplitudes no longer fully cancel at destructive-interference points). Overall, the contrast (visibility) of the fringe pattern decreases, even though the fringe spacing is unchanged.

CRQ 15.4.Why can we readily observe diffraction effects for sound wave and water waves, but not for light? Is this because light travels so much faster than these other waves? Explain.

No, it is not because of the speed of light. Diffraction becomes noticeable when the size of the gap or obstacle is comparable to, or smaller than, the wavelength of the wave. Sound waves have wavelengths of the order of centimetres to metres, and water (ripple-tank) waves have wavelengths of centimetres, both of which are comparable to the everyday gaps and obstacles (doorways, barriers) they encounter, so they diffract noticeably and spread into the "shadow" region. Visible light, however, has an extremely small wavelength (roughly 400–700 nm), which is many orders of magnitude smaller than ordinary openings and obstacles, so light appears to travel in straight lines and its diffraction is not noticeable unless it passes through a very narrow slit or fine grating whose spacing is comparable to its wavelength (as in a diffraction grating, where d is of the order of the wavelength of the light used).

CRQ 15.5.What is the significance of path difference in deriving Bragg's equation?

When a beam of X-rays strikes a crystal, it is partly reflected from successive parallel atomic planes separated by spacing d. The beam reflected from a lower plane travels an extra distance compared to the beam reflected from the plane above it; this effective path difference is 2d sinθ, where θ is the angle between the incident beam and the atomic plane. The reflected beams from all the planes can reinforce each other (constructive interference, giving a strong diffracted beam) only if this path difference equals a whole number of wavelengths, i.e., 2d sinθ = nλ. Thus the path difference is the central physical quantity that determines the condition for constructive interference; it is precisely by setting the path difference equal to an integral number of wavelengths that Bragg's equation 2d sinθ = nλ is derived, and it is this equation that allows the interplanar spacing d of a crystal to be found from the measured diffraction angle θ for X-rays of known wavelength.