

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
651. |
what type of wavefront will emerge from (i) a point source and (ii) distant light source ? |
Answer» (i) From a point source, the wavefront is diverging spherical wavefront. (ii) From a distant light source, the wavefront is plane wavefront. |
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652. |
There is a small black dot at the centre C of a solid glass sphere of refractive index `mu`. When seen from outside, the dot will appear to be locatedA. closer to the eye than its actual positionB. farther away from the eye than its actual positionC. the same as its actual positionD. independente of the refractive index of the sphere |
Answer» Correct Answer - C::D |
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653. |
In the above question, which letter will appear to be raised maximum ? |
Answer» As `mu_(v) gt mu_(r )`, therefore violet letter images will be raised maximum. | |
654. |
In the above question, how does the magnifying power change ? |
Answer» As `m = f_(0)//f_(e)`, therefore magnifying power does not change. | |
655. |
In the above question, the correct value of focal length `f` of lens as calculated from the graph is :A. `f = (1)/(OA)`B. `f = (1)/(OA + OB)`C. `f = (1)/(OA -OB)`D. `f = (1)/(OB)` |
Answer» Correct Answer - D In a convex lens, when object is at infinity, image is at focus, i.e. when `u = oo, (1)/(u) = 0`, `(1)/(v) = OB = (1)/(f) :. F = 1//OB`. |
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656. |
This question has a paragraph followed by two statements, Statement - 1 and Statement - 2. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. Statement - 1: When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of `pi`. Statement - 2 : The centre of the interference pattern is dark.A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
Answer» Correct Answer - A As light is entering from air to glass, the reflected ray suffers a phase change of `pi`. Due to it, at the centre, there will be destructive interference. Thus, both the statements are correct and statement (2) is the correct explanation of statement (1). |
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657. |
Define following terms:i. Longitudinal chromatic aberration ii. Circle of least confusion iii. Transverse chromatic aberration |
Answer» i. Longitudinal chromatic aberration: Due to different refractive indices and angle of deviations, violet and red colours of a white light converge at different focal points, fV and fR. The distance between fV and fR is measured as the longitudinal chromatic aberration. ii. Circle of least confusion: In presence of aberration the image is not a single point but always a circle. At particular location on the screen, this circle has minimum diameter. This is called circle of least confusion. iii. Transverse chromatic aberration: Radius of the circle of least confusion is called the transverse chromatic aberration. |
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658. |
Statement-1 : The dispersive power of a lens of focal length `10 cm` is `0.08`. Longitudinal chromatic aberration of the lens would be `0.8 cm`. Statement-2 : It follows from `f_( r) - f_(v) = omega.f`A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
Answer» Correct Answer - A Longitidinal chromatic aberration `(f_( r) - f_( v)) = omega f = 0.08 xx 10 = 0.8` Both the statement are true and latter is correct explanation of the former. |
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659. |
State the restrictions for having images produced by spherical mirrors to be appreciably clear. |
Answer» i. In order to obtain clear images, the formulae for image formation by mirrors or lens follow the given assumptions :
ii. In case of spherical mirrors (excluding small aperture spherical mirrors), rays farther from the principle axis do not remain parallel to the principle axis. Thus, the third assumption is not followed and the focus gradually shifts towards the pole. iii. The relation (f = R/2) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image. iv. This defect arises due to the spherical shape of the reflecting surface. |
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660. |
What are convex and concave lenses? For which condition, convex lens will have negative focal length? |
Answer» 1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive. 2. Convex lens is visualized to be internal cross section of two spheres (or one sphere or a plane surface). 3. A lens is concave if it is thicker at periphery and narrows down towards centre and has negative focal length. 4. Concave lens is visualized to be external cross section of two spheres. 5. For lenses of material optically rarer than the medium in which those are kept, convex lenses will have negative focal length and they will diverge the incident rays. |
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661. |
Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3mm. Approximately, what is the maximum distance at which these dits can be resolved by the eye? [Take wavelelngth of light =500nm]A. `1 m`B. `5 m`C. `3 m`D. `6 m` |
Answer» Correct Answer - B Limit of resolution eye `= theta = (1.22 lambda)/(D)` `= (1.22 xx 5 xx 10^(-7))/(3 xx 10^(-3)) = 2.03 xx 10^(-4) rad` If `x` is the minimum distance at width dots are just resolved, then `theta = (1 mm)/(x) = (10^(-3))/(x) = 2.03 xx 10^(-4)` `x = (10^(-3))/(2.03 xx 10^(-4)) = 5 m` |
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662. |
When a given light beam is passed through a polaroid and polaoid is rotated about the incident light axis, what are the three possiblities ? |
Answer» (i) There may be no change in intensity of emergent light, the incedent light is unpolarised. (ii) There may be change in intenstiy of emergent light with minimum not equal to zero. The incident light is partially polarised. (iii) There may be change in intensity of light with minimum equal to zero. The incident light is plane polarised. |
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663. |
A star is moving towards the earth with a speed of `4 xx 10^(7) m//s`. If the wavelength emitted by the star is `500 nm`, what would be the change in wavelength received on earth ? |
Answer» Here, `v = 4 xx 10^(7) m//s, lambda = 500 nm` `Delta lambda = ?` From `Delta lambda = -(v)/( c)lambda = -(4 xx 10^(7))/(3 xx 10^(8)) xx 500 nm` `Delta lambda = - 66.7 nm` Negative sign implies decrease in the observed wavelength. |
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664. |
At what angle `theta` above the horizon should the sun be situated so that its light reflected from the surface of still water in a pond is completely polarised. Take `mu = 1.327 and tan 53^(@) = 1.327`. |
Answer» The reflected light will be completely polarised, when `I = i_(P)`, so that `theta = (90 - i_(P))` As `mu = tan i_(P)` `1.327 = tan i_(P) : i_(P) = tan^(-1) (1.327) = 53^(@)` `:. Theta = 90^(@) - i_(P) = 90^(@) - 53^(@)` `= 37^(@)` |
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665. |
What is the angle between the plane of polariser and that of analyser to reduce the transmitted light intensity to half ? |
Answer» From Law of Malus, `I = I_(0) cos^(2)theta = (I_(0))/(2)` `:. Cos^(2) theta = (1)/(2) , cos theta = (1)/(sqrt(2)), theta = 45^(@)` |
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666. |
If the angle between the pass axis of polariser and analyser is `45^(@)`, write the ratio of intensities of original light and the transmitted light after passing through analyser. |
Answer» From Law of Malus, `I = I_(0) cos^(2) theta` `(I_(0))/(I) = (I)/(cos^(2) theta) = (1)/(cos 45^(@))^(2) = (1)/(1//sqrt(2))^(2) = 2` |
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667. |
Sound waves can not be polarized. Why ? |
Answer» Only transverse waves can be polarized. Since sound waves are longitudinal, they cannot be polrized. | |
668. |
Two polaroid `A and B` are kept in crossed position. How should a third polaroid `C` be placed between them so that the intensity of polarized light transmitted by polaroid `B` reduces to `(1)/(8)th` of the intensity of unplarised light incident on `A` ? |
Answer» Correct Answer - `45^(@)` Let the angle between the pass axis of the polaroids `A` and `C` be `theta`. As polaroids `A` and `B` are crossed, the angle between the pass axes of polaroids `C` and `B` is `(90 - theta)`. When unpolaroids light of intensity `I_(0)` falls on first polaroid `A`, intensity of polarised light through `A` is `I_(1) = I_(0)//2`. This light falls on polaroid `C` at `/_ theta`. `:.` Intensity of light transmitted through `C` is `I_(2) = I_(1) cos^(2) theta = (I_(0))/(2)cos^(2) theta` Light transmitted from `C` falls on polaroid `B` at an angle `(90 - theta)`. Therefore, intensity of light transmitted from `B` is `I_(3) = I_(2)cos^(2)(90 - theta)` `= [(I_(0))/(2)cos^(2) theta]sin^(2)theta = (I_(0))/(2)(sin theta cos theta)^(2)` `I_(3) = (I_(0))/(8)(sin 2 theta)^(2)` As `I_(3) = I_(0)//8`, therefore, therefore `(sin 2 theta)^(2) = 1` `sin 2 theta = 1, or 2 theta = 90^(@) or theta = 45^(@)` |
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669. |
You read a newspaper because of the light it reflects. Then why do you not see even a faint image of yourself in the newspaper ? |
Answer» We read newspaper because light is scttered by the paper into our eyes. This is called diffuse relfection. Newspaper has a rough surface. It cannot produce regular reflection. That is why our image cannot be seen. | |
670. |
What is Fresnel’s distance? Obtain the equation for Fresnel’s distance. |
Answer» Fresnel’s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light. Fresnel’s distance z as, z = \(\frac{a^2}{2λ}\). |
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671. |
Is the speed of light in glass independent of colour of light ? |
Answer» No, speed of light in glass depends on colour of light. From `mu = (c )/(V)`. As `mu_(v) gt mu_(r )`, therefore `V_(v) lt V_( r)` | |
672. |
Mention the differences between interference and diffraction. |
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673. |
The observed spectrum of a star shows red shift. What do you conclude ? |
Answer» Red shift in the spectrum of star indicates increase in apparent wavelength or decrease in apparent frequency of light reaching away from earth. | |
674. |
Diffraction is interference between different parts of the same wavefront. Comment. |
Answer» Yes, the statement is ture. Secondary wavelets from unobstructed part of incident wavefront interfere to produce diffraction pattern. | |
675. |
In diffraction due to a single slit, what is the condition for first minimum ? |
Answer» If `a` is width of slit, then conditon for first minimum is `a sin theta = lambda`. | |
676. |
How is resolving power of an optical instrument related to wavelength of light ? |
Answer» The resolving power of an optical intsrument varies inversely as the wavelength of light used. | |
677. |
what should be the order of size of obstacle/aperture for diffrection of light ? |
Answer» Size of obstacle/aperture should be of the same order as that of wavelength of light. | |
678. |
How are resolving power and limit of resolution of an optical instrument related ? |
Answer» Smaller is the limit of resolution, higher is the resolving power. | |
679. |
The least deflection angle of a ceratin glass prism is equal to its refracting angle. Find the latter. |
Answer» In this case we have `sin .(alpha + theta)/(2) = n sin .(theta)/(2)` (see soln. of 5.20) In our problem `alpha = theta` So, `sin theta = n sin (theta//2)` or `2 sin (theta//2) cos (theta//2) = n sin (theta//2)` Hence `cos (theta//2) = (n)/(2)` or `theta = 2 cos^(-1) (n//2) = 83^(@)`, where `n = 1.5` |
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680. |
(a) The refractive index of glass is `1.5`. What is the speed of light in glass ? (Speed of light in vaccum is `3 xx 10^(8) ms^(-1)`). (b) Is the speed of light in glass independent of colour of light ? If not, which of the two colours, red and violet travels slower in a glass prism ? |
Answer» (a) Here, `mu = 1.5, v = ?, c = 3 xx 10^(8)ms^(-1)` As `mu = ( c)/(v) :. V = ( c)/(mu) = (3 xx 10^(8))/(1.5) = 2 xx 10^(8) ms^(-1)` (b) No, the refractive index and the speed of light in a medium depend on wavelength i.e. colour of light. We know that `mu_(v) gt mu_(r )`. Therefore `v_("voilet") lt v_("red")`. Hence voilet component of white light travels slower than the red component. |
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681. |
What is the speed of light in glass of refractive index `1.5` ? Given speed of light in water is `2.25 xx 10^8 m//s` and refractive index of water is `1.3`. |
Answer» Here, `mu_(g) = 1.5, v_(g) = ? Mu_(w) = 1.3`, `v_(w) = 2.25 xx 10^(8)m//s` As `(v_(g))/(v_(w)) = (mu_(w))/(mu_(g)) = (1.3)/(1.5)` `:. v_(g) = (1.3)/(1.5)v_(w) = (1.3 xx 2.25 xx 10^(8))/(1.5)` `= 1.95 xx 10^(8)m//s` |
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682. |
What do you know about the SOFAR channel in the ocean ? |
Answer» SOFAR channel in the ocean is a particular layer of the ocean that acts like an optical fibre for sound. It allows sound to travel tens of thousands of kilomater without any loss of intensity. This channel is used by ships in distress for sending signals. it has military applications as well. The existence of this channel has been explained due to a peculiarity in the way the speed of sound changes with depth in the ocean. For details, refer to some text book on oceanography. |
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683. |
what is the cause is due colour of ocean ? |
Answer» Blue colour of ocean is due to preferential Scattering of light by water molecules. | |
684. |
The sun looks reddish at the time of sunrise and sunset. |
Answer» At the time of sunrise and sunset, the sun is near the horizon. The sun rays have to travel a larger part of atmosphere to reach the observer. As `I_(s) prop (1)/(lambda_(4))` and `lambda_(r )` is largest in the visible region, red colour is scattered the least, and enters our eyes. | |
685. |
A point source of monochromatic light emitting a lumninous flux `Phi` is positioned at the cnetre of a spherical layer of substance. The inside radius of the layer is `a`, the outside one is `b`. The coefficient of linear absorption of the substance is equal to `x`, the reflection coefficient of the surfcaes is equal to `rho`. Neglecting the secondary reflections, find the intensity of light taht passes through that layer. |
Answer» We have to derive the law of decrease of intensity in an absorbing medium taking in to account the natural geometrical fall-off (inverse sequare law) as well as absorption. Consider a thin sperical sheel of thickness `dx` and internal radius `x`. Let `I(x)` and `I(x+dx)` be the intersities at the inner and outer surfcaes of this shell. Then `4pix^(2)I(x)e^(-chi dx) = 4pi(x+dx)^(2) I(x+dx)` Except for the factor `e^(-chi dx)` this is the usual equation. we rewrite this as `x^(2)I(x) = I(x+dx) (x+dx)^(2) (1+chi dx)` `=(1+(dI)/(dx)dx) (x^(2)+2xdx)(1+chi dx)` or `x^(2)(dI)/(dx) + chi x^(2) I + 2xI = 0` Hence `(d)/(dx)(x^(2)I) + chi (x^(2)I) = 0` so `x^(2) I = Ce^(-chi x)` where `C` is a constant of intergration. In our case we apply this equation for `ale xle b` For `c le a` the usual inverse square law gives `I(a) = (Phi)/(4pia^(2))` Hence `C = (Phi)/(4pi) e^(chia)` and `I(b) = (Phi)/(4pib^(2)) e^(-chi(b -a))` This does not take into account reflection. When we do that we get `I(b) = (Phi)/(4pib^(2)) (1-rho)^(2) e^(-chi(b-a))` |
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686. |
How many times will the intensity of a narrow `X`-ray beam of wavelength `20pm` decrease after passing through a lead plate of thickness `d = 1.0 mm` if the mass absorption coefficient for the given radiation wavelength is equal to `mu//rho = 3.6 cm^(2)//g`? |
Answer» The transmission factor is `e^(-mu d)` and so the intensity will decrease `e^(mu d)` `=e^(3.6 xx 11.3 xx 0.1) = 58.4` timestimes (we have used `mu = (mu//rho) xx rho` and used the known value of density of lead). |
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687. |
A ray of light enters a rectangular glass slab of refractive index `sqrt(3)` at an angle of incidence `60^(@)`. It travels a distance of 5cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays isA. `5sqrt(3) cm`B. `5/2 cm`C. `5sqrt((3)/(2))`D. `5 cm` |
Answer» Correct Answer - B |
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688. |
A ray of light is incident at an angle of `60^(@)` on one face of a rectangular glass slab of thickness `0.1 m`, and refractive index `1.5`.Calculate the lateral shift produced. |
Answer» Correct Answer - `0.0513 m` Here, `i_(1) = 60^(@), t = 0.1m, mu = 1.5` As `(sin i_(1))/(sin r_(1)) = mu` `:. Sin r_(1) = (sin i_(1))/(mu) = (sin 60^(@))/(1.5) = 0.5773` `r_(1) = sin^(-1)(0.5773) = 35.3^(@)` Lateral shift `= (t sin (t_(1) - r_(1)))/(cos r_(1))` `= (0.1 sin (60^(@) - 35.3^(@)))/(cos 35.3^(@)) = (0.1 sin 24.7^(@))/(cos 35.3^(@))` `= (0.1 xx 0.418)/(0.816) = 0.0513 m` |
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689. |
A ray of light is incident at an angle of `45^(@)` on one face of a rectangular glass slab of thickness `10 cm` and refractive index `1.5`. Calculate the lateral shift produced. |
Answer» Here, `i_(1) = 45^(@), t = 10 cm = 0.1 m` `mu = 1.5`, lateral shift `= ?` As `mu = (sin i_(1))/(sin r_(1))` `:. sin r_(1) = (sin i_(1))/(mu) = (sin 45^(@))/(1.5) = (0.707)/(1.5) = 0.4713` `r_(1) = sin^(-1) (0.4713) = 28.14^(@)` lateral shift `= (t sin (i_(1) - r_(1)))/(cos r_(1))` `= (0.1 sin (45^(@) - 28.14^(@)))/(cos 28.14^(@))` `= (0.1 sin 16.86^(@))/(cos 28.14^(@)) = (0.1 xx 0.2900)/(0.8818) = 0.033 m` |
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690. |
A converging lens forms a real image `I` on its optic axis. A rectangular galss slab of refractive index `mu` and thickness `t` is introduced between the lens and `I`. `I` will moveA. away from the lens by `t(mu-1)`B. towards the lens by `t(mu-1)`C. away from the lens by `t(1-1//mu)`D. towards the lens by `t(1-1//mu)` |
Answer» Correct Answer - C |
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691. |
An object is placed at a distance u cm from a concave mirror of focal length f cm. The real image of the object is received on a screen placed at a distance of v cm from the mirror. The values of u are changed and the corresponding values of v are measured. Which one of the graphs shown in the figure represents the variation of `1/v` with `1/u` ?A. B. C. D. |
Answer» Correct Answer - C |
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692. |
A convex lens of focal length 40cm is in contact with a concave lens of focal length 25cm. The power of the combination isA. `+1.5`B. `-1.5`C. `+6.67`D. `-6.67` |
Answer» Correct Answer - D |
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693. |
A thin converging lens made of glass of refractive index `1.5` acts as a concave lens of focal length `50 cm`, when immersed in a liquid of refractive index `15//8`. Calculate the focal length of converging lens in air. |
Answer» Correct Answer - `20 cm` Here, `mu_(g) = 1.5, f_(l) = - 50 cm` `mu_(l) = 15//8, f_(a) = ?` `(1)/(f_(a)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` …(i) `(1)/(f_(l)) = ((mu_(g))/(mu_(l)) -1)((1)/(R_(1)) - (1)/(R_(2)))` …(ii) `(1)/(-50) = ((1.5)/(15//8) - 1)((1)/(R_(1)) - (1)/(R_(2)))` or `(1)/(R_(1)) - (1)/(R_(2)) = +(1)/(50) xx (5)/(1) = (1)/(10)` From (i), `(1)/(f_(a)) = (1.5 - 1)((1)/(10)) = (1)/(20)` `f_(a) = 20 cm` |
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694. |
A glass prism of refractive index `1.5` is immersed in water (refractive index 4/3). A light beam incident normally on the face `AB` is totally reflected to reach the face `BC`, Fig. if : .A. `sin C = 8//9`B. `sin C = 9//8`C. `sin C = 2//3`D. `sin C = 3//2` |
Answer» Correct Answer - A `(mu_(2))/(mu_(1)) = (1)/(sinC), (1.5)/(4//3) = (1)/(sin C) or sin C = (8)/(9)` |
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695. |
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m away. It is observed that the first minimum is at a distance of `2.5 mm` from the centre of the screen. Find the width of the slit. |
Answer» Here, `lambda = 500 nm = 500 xx 10^(-9)m = 5 xx 10^(-7)m, D = 1m, n = 1, x = 2.5 mm = 2.5 xx 10^(-3)m, a = ?` The condition for minima is `a(x)/(D) = n lambda` `a = (n lambda D)/(x) = (1 xx 5 xx 10^(-7) xx 1)/(2.5 xx 10^(-3)) = 2 xx 10^(-4)m = 0.2 mm` |
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696. |
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident deam. At the first maximum of the diffraction pattern the phase difference between the rays coming from the edges of the slit isA. zeroB. `pi//2`C. `pi`D. `2 pi` |
Answer» Correct Answer - C For first minimum path diff. `rarr lambda//2` phase diff. `rarr (2 pi)/(lambda) xx` path diff. `= (2 pi)/(lambda) xx (lambda)/(2) = pi` |
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697. |
In a Young’s double-slit experiment, the central bright fringe can be identified (a) as it has greater intensity than the other bright fringes (b) as it is wider than the other bright fringes (c) as it is narrower than the other bright fringes (d) by using white light instead of monochromatic light |
Answer» Correct Answer is: d) |
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698. |
What is the essential difference between fluorescence and phosphroescence ? |
Answer» Fluorescence is the phenomenon of emission of energy in visible region by a substance on absorbing radiations of higher frequency.If the process of emission continues even after the exciting radiations are stopped, the phenomenon is called phosphrescence. |
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699. |
Why does one prefer a black umbrella to a white one, even in summer ? |
Answer» A black umbrella privides a better shade than a white umbrella. Absorption of more light energy from the sun by the black unbrella does not matter because our body is not in touch with the umbrella. | |
700. |
Why does a secoundary rainbow have inverted colours ? |
Answer» This is because in the formation of a secondary rainbow, light enters from the bottom of the drop, instead of entering from the top as in case of primary rainbow and suffers two total internal reflections instead of one. | |