804 + Interview Questions in OPTICS IN GENERAL AWARENESS Page 6 InterviewSolution

251.	What are coherent sources?
Answer» Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.

Discussion

252.	How does wavefront division provide coherent sources?
Answer» Wavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.

Discussion

253.	The speed of light in an isotropic medium depends on,(a) its intensity (b) its wavelength (c) the nature of propagation (d) the motion of the source w.r.to medium
Answer» Correct answer is (b) its wavelength

Discussion

254.	How do source and images behave as coherent sources?
Answer» Source and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel’s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd’s mirror uses a source and its virtual image as two coherent sources.

254.

How do source and images behave as coherent sources?

Answer»

Source and images:

In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel’s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd’s mirror uses a source and its virtual image as two coherent sources.

Discussion

255.	What is bandwidth of interference pattern?
Answer» The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

Discussion

256.	What is diffraction?
Answer» Diffraction is bending of waves around sharp edges into the geometrically shadowed region.

Discussion

257.

Differentiate between Fresnel and Fraunhofer diffraction.

Answer»

S. No.	Fresnel diffraction	Fraunhofer diffraction
1.	Spherical or cylindrical wavefront undergoes diffraction	Plane wave front under goes diffraction
2.	Light wave is from a source at finite distance	Light wave is from a source at infinity
3.	For laboratory conditions, convex lenses need not be used	in laboratory conditions, convex lenses are to be used
4.	Difficult to observe and analyse	Easy to observe and analyse

Discussion

258.	Discuss the special cases on first minimum in Fraunhofer diffraction.
Answer» Let us consider the condition for first minimum with (n = 1). a sin θ = λ The first minimum has an angular spread of, sin θ = \(\frac{λ}{a}\). Special cases to discuss on the condition. 1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1. 2. When a ≥ λ, the diffraction is possible. For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°. For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all. 3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac{λ}{a}\)= \(\frac{λ}{2λ}\) = \(\frac{1}{2}\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

258.

Discuss the special cases on first minimum in Fraunhofer diffraction.

Answer»

Let us consider the condition for first minimum with (n = 1). a sin θ = λ

The first minimum has an angular spread of, sin θ = \(\frac{λ}{a}\). Special cases to discuss on the condition.

1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1.

2. When a ≥ λ, the diffraction is possible.

For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°.
For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.

3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac{λ}{a}\)= \(\frac{λ}{2λ}\) = \(\frac{1}{2}\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

Discussion

259.	A sphere of glass `(mu = 1.5)` is of `20 cm` diameter. A parallel beam enters it from one side. Where will it get focussed on the other side ?
Answer» Correct Answer - `5 cm`

Discussion

260.	Two lenses of focal lengths `20 cm and - 40 cm` are held in contact. The image of an object at infinity will be formed by the combination atA. `oo`B. `20cm`C. `40 cm`D. `60 cm`
Answer» Correct Answer - C

Discussion

261.	A pond of depth `20 cm` is half filled with an oil of `mu = 1.4` and the other half is filled with water of refractive index `1.33`. Calculate apparent depth of the tank when viewed normally.
Answer» Correct Answer - `14.66 cm`

Discussion

262.	A cavity of volume `V = 1.01` is filled with thernal radiation at a temperature `T = 1000K`. Find: (a) the heat capacity `C_(v)`, (b) the entropy `S` of that radiation.
Answer» (a) The total internal energy of the cavity is `U = (4 sigma)/(c )T^(4)V` Hence `C_(v) = ((delU)/(delT))_(v) = (16 sigma)/(C )T^(3)V` `= (16xx5.67xx 10^(8))/(3 xx 10^(8)) xx 10^(9) xx 10^(-3)"Joule"//^(@)K` `= (1.6 xx 5.67)/(3)nJ//K = 3.024nJ//K` (b) From first law `TdS = dU + pdV` `= VdU + UdV + (u)/(3)dV (p = (U)/(3))` `= VdU+(4U)/(3)dV` `= (16sigma)/(C)VT^(3)dT +(16sigma)/(3C)T^(4)dV` or `dS = (16sigma)/(C)VT^(2)dT +(16sigma)/(3C)T^(3)dV` `= ((16sigma)/(3C)VT^(3))` Hence `S = (16sigma)/(3C)VT^(3) = (1)/(3)C_(v) = 1.008nJ//K`.

Discussion

263.	When ray of light falls normally on a mirror, its angle of incidence is 90°. True or false? Justify your answer.
Answer» False, when light falls normally on a mirror, its angle of incidence is zero degree.

Discussion

264.	The radii of curavture of surfaces of a convex lens of glass are `20 cm` each. If `mu` is glass is `3//2`, power of lens in diopter is.
Answer» Correct Answer - `5` Here, `R_(1) = 20 cm`, `R_(2) = - 20 cm, mu = 3//2, P = ?` `(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= ((3)/(2) - 1) ((1)/(20) + (1)/(20)) = (1)/(20)` `f = 20 cm` `P = (100)/(f) = (100)/(20) = 5D`

Discussion

265.	Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
Answer» The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope. Adjustments of the spectrometer: The following adjustments must be made before doing the experiment using spectrometer. (a) Adjustment of the eyepiece: The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen. (b) Adjustment of the telescope: The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire. (c) Adjustment of the collimator: The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel. (d) Levelling the prism table: The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

265.

Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:

Answer»

The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.

Adjustments of the spectrometer: The following adjustments must be made before doing the experiment using spectrometer.

(a) Adjustment of the eyepiece: The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.

(b) Adjustment of the telescope: The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.

(c) Adjustment of the collimator: The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.

(d) Levelling the prism table: The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

Discussion

266.	Mention the types of optically active crystals with example
Answer» Types of optically active crystals: Uniaxial crystals: Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals. Biaxial crystals: Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.

266.

Mention the types of optically active crystals with example

Answer»

Types of optically active crystals:

Uniaxial crystals:

Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.

Biaxial crystals:

Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.

Discussion

267.	With what speed should a galaxy move with respect to us to that the sodium line at `589.0 nm` is observed at `589.6 nm` ?
Answer» Here, `v = ?, lambda = 589.0 nm` `Delta lambda = 589.6 - 589.0 = 0.6 nm` From `v = (Delta lambda)/(lambda) xx c = (0.6)/(589.0) xx 3 xx 10^(8) m//s`. `v = 3.06 xx 10^(5) m//s = 306 km//s` The galaxy is moving away from us, as apparent `lambda` increases.

Discussion

268.	Mona and Anushka are friends, both studying in class 12. Mona is in Science stream and Anushka in in Arts stream. Both of them go to market to purchase sunglasses. Anushka feels that any coloured glasses with fancy look are good enough. Mona tells her to look for `UV` protection glasses, polaroid glasses and photo sensitive glasses. Read the above passage and answer the following questions : (i) What are `UV` protection glasses, polaroid glasses and photo sensitive glasses ? (ii) What values are displayed by Mona ?
Answer» (i) `UV` protaction glasses are those which filter ultra-violet rays that are harmful to our eyes. Polaroid glasses help in reducing the glare. Photo-sensitive glasses get darker in strong day light. They protect our eyes from strong sunlight especially at noon. Mona has displayed concern for her friend. She has put to use the knowledge she acquired in her science classes. Mugging up things for examination is of no use. What we are taught in class room must be used in practice.

Discussion

269.	With light falling normally on a diffraction grating, the angle of diffraction of second order is equal to `45^(@)` for a wavelength `lambda_(1) = 0.65mum`. Find the angle of diffraction of third order for a wave length `lambda_(2) = 0.50 mu m`.
Answer» From the fromula `d sin theta = m lambda` we have `d sin 45^(@) =2lambda_(1) = 2 xx 0.65 mu m` or `d = 2sqrt(2) xx 0.65mu m` Then for `lambda_(2) = 0.50` in the thrid order `2sqrt(2) xx 0.65 sin theta = 3 xx 0.50` `sin theta = (1.5)/(1.3 xx sqrt(2)) = 0.81602` This gives `theta = 54.68^(@) ~~55^(@)`

Discussion

270.	Find the wavelength of monochromatic light falling normally on a diffraction grating with period `d = 2.2 mu m` if the angle between the directions to the Fraunhofer maxima of the first and the second order is equal to `Delta theta = 15^(@)`.
Answer» Given that `d sin theta_(0) = lambda` `d sin theta_(2) = d sin (theta_(1) + Delta theta) = 2lambda` Thus `sin theta_(1)cos Delta theta + cos theta_(1) sin Delta theta = 2 sin theta_(1)` or `sin theta_(1) (2- cos Delta theta) = cos theta_(1) sin Delta theta` or `tan theta_(1) = (sin Delta theta)/(2- cos Delta theta)` or `sin theta_(1) = (sin Delta theta)/(sqrt(sin^(2)Delta theta + (2-cosDelta theta)^(2)))` `= (sin Delta theta)/(sqrt(5-4 cos Delta theta))` Finally `lambda = (d sin Delta theta)/(sqrt((5 -4 cos Delta theta))`. Substituation gives `lambda = 0.534 mu m`

Discussion

271.	Assertion : In optical fibre, the diameter of the core is kept small. Reason : The small diameter of the core ensures that the fibre should have inside it an angle greater than critical angle needed for total internal reflection.A. If both, Assertion and Reason are true and the Reason is the correct explaination of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explaination of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - A Both, the Assertion and Reason are true and latter explain the former correctly. Total internal reflection of light takes place it angle of incidence is greater than critical angle. So, small diameter ensures that incident light falls at an angle greater than critical angle.

Discussion

272.	If you were driving a car, what type of mirror would you prefer to use for observing traffic at your back ?
Answer» For observing traffic at our back, we prefer to use a convex mirror. This is because a convex mirror has a much larger field of view than a plane mirror or a concave mirror. also, the nature of the image formed does not change with changing disatcne of teh object. It is always virtual, erect and smaller.

Discussion

273.	Find the free electron concentration in ionsphere if its refractive index is equal to `n = 0.90` for radiowaves of frequency `v = 100 MHz`.
Answer» From the previous problem `n^(2) = 1-(n_(0)e^(2))/(epsilon_(0) momega^(2))` `=1 - (n_(0)e^(2))/(4pi^(2)epsilon_(0)mv^(2))` Thus `n_(0) = (4pi^(2)v^(2)m epsilon_(0)//e^(2)) (1-n^(2)) = 2.36 xx 10^(7)cm^(-3)`

Discussion

274.	A motor car is fitted with a convex driving mirror of focal length `20 cm`. A second motor car is `6 m` away from the driving mirror of the first car. Calculate (i) position of second car as seen in the first car mirror. (ii) if the second car is overtaking the first car at a relative speed of `15 m//s`, how will its image be moving and in what direction ?
Answer» Correct Answer - (i) `0.1935 m` `- 1.56 cm//s` (i) Here, `f = + 20 cm = 0.2 m , u = - 6 cm` From the mirror formula `(1)/(v) = (1)/(f) - (1)/(u) = (1)/(0.2) - (1)/(-6) = (31)/(6)` `v = (6)/(31)m = 0.1935 m` As `v` is positive, image of second car is virtual, erect and is formed behind the mirror of first car. (ii) As is known, the speed of image of second car is `(dv)/(dt) = -(f^(2))/((u - f^(2))) xx (du)/(dt)` where `(du)/(dt) = 15 ms^(-1)` `:. (du)/(dt) = -((0.2)^(2))/((-6 - 0.2)^(2)) xx 15` `= - 1.56 xx 10^(2) m//s` `(dv)/(dt) = - 1.56 cm//s` Negative sign implies that as 2nd car approaches the first car, image of 2nd car moves away from convex driving mirror.

Discussion

275.	Assuming electrons of substance to be free when subjected to hard `X`-rays, determine by what magnitude the refractive index of graphite differs from unity in the case of `X-`rays whose wavelength in vaccume is equal to `lambda = 50 pm`.
Answer» For hand `x`-rays, the electrons in graphic will behave as if nearly free and the formula of previous problem can be applied. Thus `n^(2) =1 - (n_(0)e^(2))/(epsilon_(0)m omega^(2))` and `n = 1 - (n_(0)e^(2))/(2epsilon_(0)m omega^(2))` on taking square root and neglecting higher order terms. So `n-1 =- (n_(0)e^(2))/(2epsilon_(0)m omega^(2)) =- (n_(0)e^(2)lambda^(2))/(8pi^(2)epsilon_(0)m e^(2))` We calacuate `n_(0)` as follows : There are `6 xx 6.023 xx 10^(23)` electrons in `12` gms of graphite of density `1.6 gm//c.c.` Thus `n_(0) (6 xx 6.023 xx 10^(23))/((12//1.6)) per c.c` Using the values of other constants and `lambda = 50 xx 10^(-12)` metre we get `n - 1 =- 5.4 xx 10^(-7)`

Discussion

276.	Refraction of light is the phenomenon of………. When it…………. .
Answer» change in the path of light , goes from one medium to another.

Discussion

277.	An electron experiences a quasi-elastic force `kx` and `a` 'friction force' `yx` in the field of electromagnetic radiation. The `E`component of the field varies as `E = E_(0) cos omegat`. Neglecting the action of the magnetic component of the field, find, (a) the motion equation of the electron, (b) the mean power absored by the electron, the frequency at which that power is maximum and the expression for the maximum mean power.
Answer» The equation of the electron can (under the stated conditions) be written as `mddot(x) + gamma dot(x) + kx = eE_(0) cos omega t` To solve this equation we shell find it convenient to use complex displacements. Consider the equation `mddot(z) + gamma dot(z) + kz = eE_(0) e^(-i omegat)` `z = (eE_(0)e^(-i omegat))/(-m omega^(2) - i gamma omega + k)` (we igone transients) Writing `beta = (gamma)/(2m), omega_(0)^(2) = (k)/(m)` we find `z = (eE_(0))/(m)e^(-i omegat)// (omega_(0)^(2) - omega^(2) - 2i beta omega)` Now `x =`1 Real part of `z` `= (eE_(0))/(m). (cos (omegat + varphi))/(sqrt((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2))) = a cos (omega t + varphi)` where `tan varphi = (2 beta omega)/(omega^(2) - omega_(0)^(2))` `(sin varphi =- (2 beta omega)/(sqrt((omega_(0)^(2) - omega_(0)^(2))^(2) + 4beta^(2) omega^(2))))` (b) We calculate the power absorded as `P = lt Edot(x) gt = lt eE_(0)cos omegat (-omega a sin (omegat + varphi)) gt` `=eE_(0).(eE_(0))/(m)(1)/(2). (2 beta omega)/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2)).omega = ((eE_(0))/(m))^(2) (beta m omega^(2))/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2) omega^(2))` This is clearly maximum when `omega_(0) = omega` because `P` can be written as `P = ((eE_(0))/(m))^(2) (beta omega)/((omega_(0)^(2)/ omega^(2)-omega)^(2) + 4beta^(2))` and `P_(max) = (m)/(4 beta) ((eE_(0))/(m))^(2)` for `omega = omega_(0)`. `P` can also be calculated from `P = lt gamma dot(x). dot(x) gt` `= (gamma omega^(2) a^(2)//2) = (beta m omega^(2) (eE_(0)//m)^(2))/((omega_(0)^(2) - omega^(2))^(2) + 4beta^(2)omega^(2))`.

Discussion

278.	The basic cause of refraction is…………in going…………. .
Answer» change in the velocity of light , from one medium to another.

Discussion

279.	Does the magnifying power of a microscope depend on colour of light used ? Justify your answer.
Answer» Magnifying power of a compound microscope is `M = (L)/(f_(0))(1 + (1)/(f_(e)))` As `f_(0)` and `f_(e)` both depend on colour of light, therfore, amgnifying power of a microscope upon colour of light.

Discussion

280.	(a) List some advantanges of a reflecting telescope, especially for high resolution astronomy. (b) A reflecting type telescope has a large mirror for its objective with radius of curvature equal to `80 cm`. What is the magnifying power of telescope if eye piece used has a focal length of `1.6 cm` ?
Answer» (a) (i) There is no chromatic aberration as the objective is a mirror. (ii) Spherical aberration is reduced using mirror objective in the form of a paraboloid. (iii) Image is brighter compared to that in a refractine telescope. (iv) Mirror requires griding and polishing of only one side. (v) High resolution is resolution is achieved by using a mirror of large aperture (b) Here, `R = 80 cm`. `f_(0) = (R )/(2) = (80)/(2) = 40 cm` and `f_(e) = 1.6 cm`. As magnifying power, `m = (f_(0))/(f_(e))` `:. m = (40)/(1.6) = 25`

Discussion

281.	Does critical angle depend on colour of light ?
Answer» Correct Answer - Yes, it depends.

Discussion

282.	Hoe does magnifying power change with change in length of tube for a given microscope ?
Answer» In case of a microscope, magnifying power `\|m\| = (v_(0))/(u_(0)) xx (d)/(u_(e))`, where `L = v_(0) + u_(e)` When `L` is increased, `u_(e)` increases as `v_(0)` is fixed `:. \|m\|` will decrease.

Discussion

283.	In single slit diffraction pattern, how does the angular separation between firnges change when distance of screen from slit is doubled ?
Answer» Angular separation `= (beta)/(D) = (lambda)/(d)` as both, `lambda and d` remain unchanged, angular separation between fringes remains the same.

Discussion

284.	What is the essential conditon for diffraction of light ?
Answer» Diffraction of light occurs only when size of obstecle/aperture is of the order of wavelength of light.

Discussion

285.	A girl is using speces of `f = -50 cm`. Name the defect of her vision and calculate power of lens to be used.
Answer» As `f` nagative, the lens is concave. The girl is suffering from myopia, `P = (100)/(f) = (100)/(- 50) = - 2D`

Discussion

286.	The diameter of the sun is `~~ 10^9 m`, but it appears to be a small disc, why ?
Answer» This is because the distance of the sun from the earth is very large `(~~ 10^(11)m)`. `:.` Angle subtended by the sun at our eye `~~ 10^(9)//10^(11) = 10^(-2)` radian. The same angle is subtended by a disc of diameter `1 cm` at a disatnce of `100 cm`. From the eye. Hence the sun looks like a small disc.

Discussion

287.	a convex lens forms a virtual image of an object. What is the position of the object ?
Answer» The object lies between optical centre and principal focus of the lens.

Discussion

288.	An object is placed at the focus of a concave lens. Where will be image ?
Answer» The image will be formed at infinity. (Use this formula, taking `u = - f`).

Discussion

289.	Mr. Chawla and Mr. Batra are two friends. Both are senior citizens. Mr. Batra once fell-ill. He took medicine from a quack who diagnonses by pulse check only. The medicine did not work and he continued ailing for along. Mr. Chawla then took him to a qualified doctor who diagnosed the cause of his illness by testing his blood sample and urine sample using clinical microscopes. The medicine prescribed by the doctor worked and Mr. Batra got well within a couple of days. Read the above passage and answer the following questions : (i) Should we go in for clinical tests ? Why ? (ii) Should we go to specialists inspite of their exorbitant charges ? (iii) What values did Mr. Chawla display ?
Answer» (i) Yes, we must go in for clinical tests as recommended by attending physician. These tests are necessary to ascertain the exact cause of ailment. (ii) Healthy life is more precious than money. Therefore, we should consult specialists as and when required, to take advantage of the repaid advances in medical sciences. (iii) Mr. Chawla displays health concern for his friend and advised him to take right course of action.

Discussion

290.	Consider an extended object immersed in water contained in a plane through. When seen from close to the edge of the through, the object looks distorted because.A. the apparent depth of the points close the egde are nearer the surface of the water compared to the points away from the edge.B. the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.C. some of the points of the object far away from the edge may not be visible because of total internal reflection.D. water in a trough acts as a lens and magnifies the object.
Answer» Correct Answer - A::B::C Here, an extended object lies immersed in water contained in a plane trough. When seen from close to the edge of the trough, the object looks distorted on account of refraction of light from denser to rarer medium, Therefore, apparent depths of the points close to the egde and nearer to the surface of water is more compared to points away from the edge. Further, the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Again, some of the points of the object, far away from the edge may not be visible because of total internal reflection. The choices (a), (b), (c ) are correct.

Discussion

291.	If VB, VG, VR be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?(a) VB = VG = VR(b) VB > VG > VR(c) VB < VG < VR(d) VB < VG >VR
Answer» (c) V_B < V_G < V_R

Discussion

292.	List any five properties of light?
Answer» Light is a form of energy. Light always travels along a straight line. Light does not need any medium for its propagation. It can even travel through a vacuum. The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1 Since light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = vλ (c = velocity of light). Different coloured light has a different wavelength and frequency.

Discussion

293.	The scattering of light by colloidal particles in the colloidal solution is called: (a) Raman scattering (b) Tyndall scattering (c) Mie scattering (d) Elastic scattering
Answer» (b) Tyndall scattering

Discussion

294.	Can a convex mirror form a mgnified image ?
Answer» Correct Answer - No, it cannot.

Discussion

295.	Power `(P)` of a lens is given by reciprocal of focal length `(f)` of the lens. i.e. `P = 1//f`. When `f` is in metre, `P` is in dioptre. For a convex lens, power is positive and for a concave lens, power is negative. When a number of thin lenses of powers `p_(1), p_(2), p_(3)....` are held in contact with one another, the power of the combination is given by algebraic sum of the powers of all the lenses i.e., `P = p_(1) + p_(2) + p_(3) + ........` Answer the following questions : Two thin lenses are in contact and the focal length of the combination is `80 cm`. If the focal length of one lens is `20 cm`, the focal length of the other would beA. `-26.7 cm`B. `60 cm`C. `80 cm`D. `20 cm`
Answer» Correct Answer - A Here, `f_(1) = 20 cm. f_(2) = ? F = 80 cm`. As `(1)/(f_(1)) + (1)/(f_(2)) = (1)/(F)` `:. (1)/(f_(2)) = (1)/(F) - (1)/(f_(2)) = (1)/(80) - (1)/(20) = -(3)/(80)` `f_(2) = -(80)/(3) = - 26.7 cm`

Discussion

296.	A polariod (I) is placed in front of a monochromatic source. Another polariod (II) is placed in front of this polariod (I) and rotated till no light passes. A third polariod (III) is now placed in between (I) and (II), then
Answer» As per the given question, monochromatic light emerging from polaroid (I) is plane polarized. When polaroid (II) is placed in front of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at `90^(@)`. When a third polaroid (III) is placed inbetween (I) and (II). In all other cases, light will emerge from (II), as pass axis of (II) will no longer be at `90^(@)` to the pass axis of (III).

Discussion

297.	A point source of light with wavelength `lambda = 0.50mu m` is located at a disatnce `a = 100cm` in front of a diaphragm with round aperture of radius `r = 1.0 mm`. Find the distance `b` between the diaphragm and the observation point for which the number of Fresnel zones in the aperture equals `k = 3`.
Answer» By definition `r_(k)^(2) = k(ab lambda)/(a + b)` for the periphery of the `K^(th)` zone. Then `ar_(k)^(2) + br_(k)^(2) = kab lambda` So `b = (ar_(k)^(2))/(ka lambda - r_(k)^(2)) = (ar^(2))/(ka lambda - r^(2)) = 2` metre. on putting the values.(It is given that `r = r_(k))` for `k = 3`.

Discussion

298.	Why does a ray of light bend towards normal as it passes from air to glass ?
Answer» As `mu = (sin i)/(sin r) = (c )/(v)`, speed of light in glass `(v)` lt speed of light in air `(c )` `:. sin r lt sin i`. Hence a ray of light bends towards normal as it passes from air to glass.

Discussion

299.	Can the relative index of a medium w.r.t. another medium be less than unity ?
Answer» Yes, for example, the relative refractive index of water w.r.t. glass `= (4//3)/(3//2) = (8)/(9) lt 1`.

Discussion

300.

The refractive indices of the material of the prism for red and yellow colour are 1.620 and 1.635 respectively. Calculate the angular dispersion and dispersive power, if refracting angle is 8°.

Answer»

Given: n_R = 1.620, n_Y = 1.635, A = 8°

To find:

i. Angular dispersion (δ_V – δ_R)

ii. Dispersive power (ω)

Formulae:

i. δ_v – δ_r = A(n_V – n_R)

ii. ω = \(\frac{n_V-n_R}{n_Y-1}\)

Calculation: Since, n_Y = \(\frac{n_V+n_R}{2}\)

∴ n_V = 2n_Y – n_R

n_V = 2 × 1.635 – 1.620 = 3.27 – 1.620

∴ n_V = 1.65

From formula (i),

δ_V – δ_R = 8(1.65 – 1.620)

= 8 × 0.03 = 0.24°

∴ δ_V – δ_R = 0.24°

From formula (ii),

ω = \(\frac{1.65-1.620}{1.635-1}=\frac{0.03}{0.635}\) = 0.0472

Discussion

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