680 + Interview Questions in WAVE-OPTICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength `6000Å`. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will beA. `3500 Å`B. `4200 Å`C. `4700 Å`D. `6000 Å`
Answer» Correct Answer - B

Discussion

2.	Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.
Answer» `becausesintheta=(lamda)/(a)" "theta=` half angular width of the central maximum `a=12xx10^(-5)cm,lamda=6000Å=6xx10^(-5)cm` `thereforesintheta=(lamda)/(a)=(6xx10^(-5))/(12xx10^(-5))=0.50impliestheta=30^(@)`

Discussion

3.	Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.A. `40^(@)`B. `45^(@)`C. `30^(@)`D. `60^(@)`
Answer» Correct Answer - C

Discussion

4.	In a single slit diffraction, the width of slits is `0.5cm`, focal length of lens is 40 cm and of first dark fringe is
Answer» Correct Answer - B `a sin theta = n lambda a.(Y)/(D)= n lambda , But = D=f` `a. (Y)/(D) = n lambda Y= (n lambda f)/(a)`

Discussion

5.	What will be the angular width of central maxima in Fraunhofer diffraction when light of wavelength `6000Å` is used and slit width is `12xx10^-5cm`?A. (a) `2rad`B. (b) `3rad`C. (c) `1rad`D. (d) `8rad`
Answer» Correct Answer - C Angular width `=(2lambda)/(d)=(2xx6000xx10^(-10))/(12xx10^-5xx10^-2)=1rad.`

Discussion

6.	A light wave is incident normally over a slit of width `24xx10^-5cm`. The angular position of second dark fringe from the central maxima is `30^@`. What is the wavelength of light?A. `6000 Å`B. `5000 Å`C. `3000 Å`D. `1500 Å`
Answer» Correct Answer - A For nth dark fringe, in single slit `a sin theta=nlambda` For second dark fringe, `a sin theta=2lambda` `So, 24xx10^(-5)xx10^(-2)xxsin 30^(@)=2lambda` `:. lambda=6xx10^(-7)m=6000Å`

Discussion

7.	Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.9^(@)`B. `0.18^(@)`C. `0.54^(@)`D. `0.36^(@)`
Answer» Correct Answer - B

Discussion

8.	Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.36^(@)`B. `0.18^(@)`C. `0.72^(@)`D. `0.09^(@)`
Answer» Correct Answer - A `d sin theta=lambda rArr theta approx lambda/d` `:. 2theta=(2lambda)/d=(2xx6328xx10^(-10))/(0.2xx10^(-3))xx180^(@)/pi=0.36^(@)`

Discussion

9.	Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?A. `12 xx 10^(-5)cm`B. `18xx10^(-5)cm`C. `24xx10^(-5)cm`D. `36xx10^(-5)cm`
Answer» Correct Answer - C

Discussion

10.	Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?
Answer» `asintheta=2lamda` given `lamda=6xx10^(-7)m,a=24xx10^(-5)xx10^(-2)`m `sintheta=(2lamda)/(a)=(2xx6xx10^(-7))/(24xx10^(-7))=(1)/(2)" "thereforetheta=30^(@)`

Discussion

11.	A parallel beam of wavelength `lambda=450xx10^9m` passes through a long slit of width `2xx10^-4m`. The angular divergence for which most of light is diffracted isA. `(2pi)/(3)`B. `(5pi)/(4)`C. `(3pi)/(4)`D. `(pi)/(3)`
Answer» Correct Answer - B `a sin theta = n lambda , theta = (lambda)/(a) (n pi)/(180)rad`

Discussion

12.	Light of wavelength `6000 Å` from a distance source falls on a slit `0.5` mm wide. The distance between two dark bands on each side of the central bright band of the diffraction pattern observed on a screen placed at a distance 2m from the slit isA. `1.2nm`B. `2.4nm`C. `3.6nm`D. `4.8mm`
Answer» Correct Answer - D Required distance `= (2D lambda)/(d)`

Discussion

13.	Among two interfering sources, let `A` be ahead in phase by `54^(@)` relative to `B`. If the observations be taken from point `P`, such that `PB - PA = 1.5 lambda`, deduce the phase difference between the waves from `A` and `B` reaching `P`.
Answer» Correct Answer - `3.3 pi`

Discussion

14.	Light of wavelength `6000 A^(@)` is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the silt. The slit width will beA. `0.3mm`B. `0.2mm`C. `0.15mm`D. `0.1mm`
Answer» Correct Answer - A From, `a sin theta = n lambda` , `a(x)/(D) = n lambda` or `a = (n lambda D)/(x) = (1xx6000xx10^(-10)xx2)/(4xx10^(-3))` `:. A = 3xx10^(-4)= 0.3mm`

Discussion

15.	A light of wavelength, 5000Å is coming from a distant star. What is the limit of resolution of a telescope whose objective has a diameter of 200 cm ?
Answer» The limit of resolution `Deltatheta` is given by `Delta theta~~(0.61lambda)/a` Here `lambda=5000Å=5000xx10^(-10)m` `a=200/2=100 cm =100xx10^(-2)m` putting the values we have `Deltatheta~~3.05xx10^(-7)` radians

Discussion

16.	A telescope is used to resolve two stars separated by `4.6xx10^(-6)` rad. If the wavelength of light used is `5460 Å` , what should be the aperture of the objective of the telescope ?A. 0.1488 mB. 0.567 mC. 1 mD. 2 m
Answer» Correct Answer - A Aperture of the telescope, `D=(1.22lambda)/(d theta)` Here, `lambda=5460Ã…=5460xx10^(-10)m, d theta=4.6xx10^(-6)`rad, ` :. D=(1.22 xx 5460xx10^(-10))/(4.6xx10^(-6))=0.1488` m

Discussion

17.	Two stars are situated at a distance of 8 light years from the earth. These are to be just resolved by a telescope of diameter 0.25 m. If the wavelength of light used is 5000 Å, then the distance between the stars must beA. `3 xx 10^(10)m`B. `3.35 xx 10^(11) m`C. `1.95 xx 10^(11)m`D. `4.32 xx 10^(10)m`
Answer» Correct Answer - D

Discussion

18.	A telescope is used to resolve two stars separated by `4.6xx10^(-6)` rad. If the wavelength of light used is `5460 Å` , what should be the aperture of the objective of the telescope ?A. `0.448m`B. `0.1448m`C. `1.1448m`D. `0.011m`
Answer» Correct Answer - B The aperture (a) of the telescope is given as `a = (1.22lambda)/(theta)`

Discussion

19.	Angular width `(beta)` of central maximum of a diffraction pattern on a single slit does not depend uponA. `lambda` of light usedB. Width of slitC. Distance of slits from screenD. Ratio of `lambda` and slit width
Answer» Correct Answer - A

Discussion

20.	What will be the angle of diffracting for the first minimum due to Fraunhofer diffraction with sources of light of wavelength `550nm` and slit of width `0.55mm`?A. (a) `0.001rad`B. (b) `0.01rad`C. (c) `1rad`D. (d) `0.1rad`
Answer» Correct Answer - A Using `dsin theta=nlambda`, for `n=1` `sin theta=lambda/d=(550xx10^-9)/(0.55xx10^-3)=10^-3=0.001rad`

Discussion

21.	What is the approximate radius of the central bright differaction spot of light of wavelength `lambda=0.5 mum`, if focal length of the lens is 20 cm and radius of aperture of the lens is 5 cm ?
Answer» `r_(0)~~(0.61 lambdaf)/a` Putting values `r_(0)=(0.61xx0.5xx10^(-6)xx20xx10^(-2))/(5xx10^(-2))=1.22xx10^(-6)m=1.22 mum`

Discussion

22.	In a single slit diffraction pattern obtained on a screen, if the slit width is gradually decreased, which of the following statement is/are correct:A. Fringe pattern will expandB. Fringe pattern will shrinkC. Fringes may disappearD. No effect on fringe pattern
Answer» Correct Answer - A

Discussion

23.	A plane wave of wavelength `6250 Å` is incident normally of the principal maximum on a screen distant 50 cm will beA. `312.5 xx 10^(-2)cm`B. `312.5xx10^(-4)cm`C. `312cm`D. `312.5xx10^(-5)cm`
Answer» Correct Answer - A Here, `lambda = 6250 Å = 6520 x 10^(-10)m` `a = 2xx10^(-2)cm = 2xx10^(-4)m` `D = 50 cm = 0.5 m` `:.` Width of central maxima `= (2lambda D)/(a)` `= (2xx6250xx10^(-10)xx0.5)/(2xx10^(-4)) = 312.5xx10^(-3) cm`

Discussion

24.	The diameter of an objective of a telescope, which can just resolve two stars situated at angular displacement of `10^(-4)` degee, should be `(lambda = 5000 Å)`A. `35mm`B. `35cm`C. `35m`D. `3.5cm`
Answer» Correct Answer - B `theta = (1.22lambda)/(a)`

Discussion

25.	Angular width `(beta)` of central maximum of a diffraction pattern on a single slit does not depend uponA. (a) Distance between slit and sourceB. (b) Wavelength of the slitC. (c) Width of the slitD. (d) Frequency of light used
Answer» Correct Answer - A For single slit diffraction pattern `d sin theta=lambda(d=slit wi dth)` Angular width `=2theta=2sin^-1(lambda/d)` It is independent of D, i.e., distance between screen and slit.

Discussion

26.	In Fresnel diffraction, if the distance between the disc and the screen is decreased, the intensity of central bright spot willA. (a) IncreaseB. (b) DecreaseC. (c) Remain constantD. (d) None of these
Answer» Correct Answer - B `A=npidlambdaimpliesnd=(A)/(pilambda)=const ant` `impliesnprop1/d`(n=number of blocked HPZ) on decreasing d, n increases, hence intensity decreases.

Discussion

27.	The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 Å. Then the slit width will beA. `5mm`B. `2.5mm`C. `1.25mm`D. `1.0mm`
Answer» Correct Answer - B Distance between first and sixth minima `x = (5lambda D)/(a)`

Discussion

28.	The two slits are 1 mm apart from each other and illuminated with a light of wavelength `5xx10^(-7)` m. If the distance of the screen is 1 m from the slits, then the distance between third dark fringe and fifth bright fringe isA. 1.2 mmB. 0.75 mmC. 1.25 mmD. 0.625 mm
Answer» Correct Answer - C Given : `lambda=5xx10^(-7)m, D=1 m, d=1 mm.` Distance of `n^(th)` bright fringe from the centre `= (nD lambda)/(d)` where `n=1,2,3, â€¦â€¦` So the distance fo `5^(th)` bright fringe `= (5D lambda)/(d)` Distance of `n^(th)` dark fringe from the centre `=(n-(1)/(2))(D lambda)/(d)` where `n=1,2,3,4, .........` `3^(rd)` dark fringe `=(3-(1)/(2))(D lambda)/(d)=(5)/(2)(D lambda)/(d)` Distance between them `=(5-(5)/(2))(D lambda)/(d)=(5)/(2)(D lambda)/(d)` `=(5xx1xx5xx10^(-7))/(2xx1xx10^(-3))=12.5xx10^(-4)m=1.25mm.`

Discussion

29.	A single slit of width `0.20mm` is illuminated with light of wavelength `500nm`. The observing screen is placed `80cm` from the slit. The width of the central bright fringe will beA. (a) `1mm`B. (b) `2mm`C. (c) `4mm`D. (d) `5mm`
Answer» Correct Answer - C Width of central bright fringe. `=(2lambdaD)/(d)=(2xx500xx10^-9xx80xx10^-2)/(0.20xx10^-3)` `=4xx10^-3m=4mm`.

Discussion

30.	A parallel beam of light of wavelength `6000Å` gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :A. `6 xx 10^(-3)` radB. `1.8 xx 10^(-3)` radC. `3 xx 10^(-3)` radD. `2 xx 10^(-3)` rad
Answer» Correct Answer - A

Discussion

31.	Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle `theta` . The phase difference between the ray reflected by the top surface of the glass and the bottom surface isA. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+pi`B. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)`C. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+(pi)/(2)`D. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+2pi`
Answer» Correct Answer - A

Discussion

32.	In single slit fraunhoffer diffraction which type of wavefront is required-A. cylindricalB. sphericalC. ellipticalD. plane
Answer» Correct Answer - D

Discussion

33.	A ray of light is incident on the left vertical face of the glass slab. If the incident light has an intensity I and on each reflection the intensity decreases by 90% and on each refraction the intensity decreases by 10%, find the ratio of the intensities of maximum to minimum in reflected pattern.
Answer» Correct Answer - [361]

Discussion

34.	The wavefront of a light beam is given by the equation `x + 2y + 3x = c` (where c is arbitrary constant), then the angle made by the direction of light with the y-axis isA. `cos^(-1).(1)/(sqrt(14))`B. `sin^(-1).(2)/(sqrt(14))`C. `cos^(-1).(2)/(sqrt(14))`D. `sin^(-1).(3)/(sqrt(14))`
Answer» Correct Answer - C Here direction of light is given by normal vector `vec(n)=hati+2hatj+3hatk` `:.` angle made by the `vec(n)` with y-axis is given by `cos beta= 2/(sqrt(1^(2)+2^(2)+3^(2)))=2/(sqrt(14))`

Discussion

35.	A beam of light travelling in water falls on a glass plate immersed in water. When the incident angle is `51^(@)`, the reflected beam of light is found to be completely plane polarised. Determine the refractive index of glass. Given refractive idex of water =4/3.
Answer» Correct Answer - 1.647

Discussion

36.	Sodium lamps are used in foggy conditions becauseA. (a) yellow light is scattered more by the fog particlesB. (b) yellow light is unaffected during its passage through the fogC. (c) yellow light is scattered less by the fog particlesD. (d) wavelength of yellow light is the mean of the visible part of the spectrum
Answer» Correct Answer - C Sodium light emits monochromatic light which has only one colour (deep yellow). This colour is scattered less by the fog particles. While the other light sources comprise lights of many different colours (of different wavelengths) so they are scattered most due to they provide some degree of colour rendering.

Discussion

37.	When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light isA. (a) polarisedB. (b) of longer wavelengthC. (c) of shorter wavelengthD. (d) of high intensity
Answer» Correct Answer - C When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light is of shorter wavelength, because `Ac curacyprop(1)/(Wavel eng th)`

Discussion

38.	A double slit experiment is performed with light of wavelength `500nm`. A thin film of thickness `2mum` and refractive index `1.5` is introduced in the path of the upper beam. The location of the central maximum willA. (a) Remain unshiftedB. (b) shift downward by nearly two fringesC. (c) shift upward by nearly two fringesD. (d) shift downward by 10 fringes
Answer» Correct Answer - C If shift is equal to n fringes width, then `n=((mu-1)t)/(lambda)=((1.5-1)xx2xx10^-6)/(500xx10^-9)` `=(1)/(500)xx10^3=2` Since a thin film is introduced in upper beam. So shift will be upward.

Discussion

39.	A monochromatic beam of light fall on YDSE apparatus at some angle (say `theta`) as shown in figure. A thin sheet of glass is inserted in front of the lower slit `s_(2)`. The central bright fringe (path difference `= 0`) will be obtained A. (a) at OB. (b) Above OC. (c) Below OD. (d) Anywhere depending on angle `theta`, thickness of plate t and refractive index of glass `mu`
Answer» Correct Answer - D If `dsintheta=(mu-1)t`, central fringe is obtained at O If `dsinthetagt(mu-1)t`, central fringe is obtained above O and If `dsinthetalt(mu-1)t`, central fringe is obtained below O.

Discussion

40.	A monochromatic beam of light fall on YDSE apparatus at some angle (say `theta`) as shown in figure. A thin sheet of glass is inserted in front of the lower slit `s_(2)`. The central bright fringe (path difference `= 0`) will be obtained A. (a) At OB. (b) Above OC. (c) Below OD. (d) Anywhere depending on angle `theta`, thickness of plate t and refractive index of glass `mu`
Answer» Correct Answer - D If `dsintheta=(mu-1)t`, central fringe is obtained at O If `dsinthetagt(mu-1)t`, central fringe is obtained above O and If `dsinthetalt(mu-1)t`, central fringe is obtained below O.

Discussion

41.	In YDSE, `d=5lambda`, then the total no. of maxima observed upon screen will beA. 9B. 8C. 7D. 5
Answer» Correct Answer - A

Discussion

42.	Statement-1: In YDSE central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen. Statement-2: In an interference pattern, whatever energy disappears at the minimum, appear at the maximum.A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1B. Statement-2 is true, statement-2 is true: statement-2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true.
Answer» Correct Answer - B

Discussion

43.	A double slit experiment is performed with light of wavelength `500nm`. A thin film of thickness `2mum` and refractive index `1.5` is introduced in the path of the upper beam. The location of the central maximum willA. reamain unshiftedB. shift downward by nearly two fringesC. shift upward by nearly two fringesD. shift downward by 10 fringes
Answer» Correct Answer - C `(mu-1)t=n lambda`

Discussion

44.	A monochromatic beam of light fall on YDSE apparatus at some angle (say `theta`) as shown in figure. A thin sheet of glass is inserted in front of the lower slit `s_(2)`. The central bright fringe (path difference `= 0`) will be obtained A. at OB. above OC. below OD. anywhere depending on angle `theta` thickness of plate t and refractive index of glass `mu`
Answer» Correct Answer - A

Discussion

45.	In double slit experiment fringes are obtained using light of wavelength `4800 Å` One slit is covered with a thin glass film of refractive index. `1.4` and another slit is covered by a film of same thickness but refractive index `1.7`. By doing so, the central fringe is shifted to fifth bright fringe in the original pattern. The thickness of glass film isA. `2xx10^(-3)mm`B. `4xx10^(-3)mm`C. `6xx10^(-3)mm`D. `8xx10^(-3)mm`
Answer» Correct Answer - D `n lambda= (mu_(2)-mu_(1))t, 5beta= (mu_(2)-mu_(1))t(beta)/(lambda)`

Discussion

46.	A material having an index of refraction of 1.30 is used as an antireflective coating on a piece of glass `(n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 500 nm light?
Answer» Treating the anti-reflectance coating like a camera lens coating, `2 t = (m + (1)/(2)) (lambda)/(n)` Let `m = 0: t = (lambda)/(4n) = (3.00cm)/(4(1.50))= 0.500 cm` This anti-reflectance coating could be easily countered by changing the wavelength of the radar to 1.50 cm - now creating maximum reflection.

Discussion

47.	Lights of wavelengths `lambda_(1)=4500 Å, lambda_(2)=6000 Å` are sent through a double slit arrangement simultaneously. ThenA. no interference pattern will be formedB. the third order bright fringe of `lambda_(1)` will coincide with the fourth order bright fringe of `lambda_(2)`C. the third order bright fringe of `lambda_(2)` will coincide with the fourth order bright fringe of `lambda_(1)`D. the fringes of wavelength `lambda_(1)` will be wider than the fringes of wavelength `lambda_(2)`
Answer» Correct Answer - C

Discussion

48.	Monochromatic light of wavelength `5000 Å` is used in YDSE, with slit width, `d = 1 mm`, distance between screen and slits, `D = 1 M`. If intensites at the two slits are `I_1 = 4I_0 and I_2 = I_0`, find: a. finge width `beta:` b. distance of 5th minima from the central maxima on the screen, c. intensity at `y = (1)/(3) mm,` d. distance of the 1000th maxima, and e. distance of the 5000th maxima.
Answer» (i) `beta=(lambdaD)/(d) =(5000xx10^(-10)xx1)/(1xx10^(-3))=0.5 mm` (ii) `y=(2n-1) (lambdaD)/(2d),n=5 rArr y=2.25 mm` (iii) At `y=(1)/(3) mm, y lt lt D` Hence `Deltap=(dy)/(D)` `Delta phi=(2pi)/(lambda)Deltap=2pi (dy)/(lambdaD)=(4pi)/(3)` Now resultant intensity `I= I_(1)+I_(2) sqrt(I_(1)I_(2)) cos Delta phi =4I_(0)+I_(0)+2 sqrt(4I_(0)^(2)) cos Delta phi =5I_(0)+4I_(0) cos (4pi)/(3) =3I_(0)` (iv) `(d)/(lambda)=(10^(-3))/(0.5xx10^(-6))=2000` `n=1000` is not ` lt lt 2000` Hence now `Deltap=d sin theta` must be used Hence, `d sin theta =nlambda=1000lambda` `rArr sin theta=1000(lambda)/(d)=(1)/(2) rArr theta =30^(@)` `y=D tan theta=(1)/(sqrt3)` meter (v) Higest order maxima `n_(max)=[(d)/(lambda)]=2000` Hence, `n=5000` is not possible.

Discussion

49.	If yellow light of `5890 Å` is incident on a grating that has 850 lines per mm. what will be the angle between both the first order images?
Answer» Correct Answer - `60^(@)`

Discussion

50.	With a thin air film between two points, 6 fringes appear when light of wavelength 5890 Å is used. Calculate the difference in the thickness of the film between the two points.
Answer» Correct Answer - 1.767 micron

Discussion

Explore topic-wise InterviewSolutions in .

Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.

Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.A. `40^(@)`B. `45^(@)`C. `30^(@)`D. `60^(@)`

In a single slit diffraction, the width of slits is `0.5cm`, focal length of lens is 40 cm and of first dark fringe is

What will be the angular width of central maxima in Fraunhofer diffraction when light of wavelength `6000Å` is used and slit width is `12xx10^-5cm`?A. (a) `2rad`B. (b) `3rad`C. (c) `1rad`D. (d) `8rad`

A light wave is incident normally over a slit of width `24xx10^-5cm`. The angular position of second dark fringe from the central maxima is `30^@`. What is the wavelength of light?A. `6000 Å`B. `5000 Å`C. `3000 Å`D. `1500 Å`

Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.9^(@)`B. `0.18^(@)`C. `0.54^(@)`D. `0.36^(@)`

Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :A. `0.36^(@)`B. `0.18^(@)`C. `0.72^(@)`D. `0.09^(@)`

Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?A. `12 xx 10^(-5)cm`B. `18xx10^(-5)cm`C. `24xx10^(-5)cm`D. `36xx10^(-5)cm`

Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?

A parallel beam of wavelength `lambda=450xx10^9m` passes through a long slit of width `2xx10^-4m`. The angular divergence for which most of light is diffracted isA. `(2pi)/(3)`B. `(5pi)/(4)`C. `(3pi)/(4)`D. `(pi)/(3)`

Light of wavelength `6000 Å` from a distance source falls on a slit `0.5` mm wide. The distance between two dark bands on each side of the central bright band of the diffraction pattern observed on a screen placed at a distance 2m from the slit isA. `1.2nm`B. `2.4nm`C. `3.6nm`D. `4.8mm`

Among two interfering sources, let `A` be ahead in phase by `54^(@)` relative to `B`. If the observations be taken from point `P`, such that `PB - PA = 1.5 lambda`, deduce the phase difference between the waves from `A` and `B` reaching `P`.

Light of wavelength `6000 A^(@)` is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the silt. The slit width will beA. `0.3mm`B. `0.2mm`C. `0.15mm`D. `0.1mm`

A light of wavelength, 5000Å is coming from a distant star. What is the limit of resolution of a telescope whose objective has a diameter of 200 cm ?

A telescope is used to resolve two stars separated by `4.6xx10^(-6)` rad. If the wavelength of light used is `5460 Å` , what should be the aperture of the objective of the telescope ?A. 0.1488 mB. 0.567 mC. 1 mD. 2 m

A telescope is used to resolve two stars separated by `4.6xx10^(-6)` rad. If the wavelength of light used is `5460 Å` , what should be the aperture of the objective of the telescope ?A. `0.448m`B. `0.1448m`C. `1.1448m`D. `0.011m`

Angular width `(beta)` of central maximum of a diffraction pattern on a single slit does not depend uponA. `lambda` of light usedB. Width of slitC. Distance of slits from screenD. Ratio of `lambda` and slit width

What will be the angle of diffracting for the first minimum due to Fraunhofer diffraction with sources of light of wavelength `550nm` and slit of width `0.55mm`?A. (a) `0.001rad`B. (b) `0.01rad`C. (c) `1rad`D. (d) `0.1rad`

What is the approximate radius of the central bright differaction spot of light of wavelength `lambda=0.5 mum`, if focal length of the lens is 20 cm and radius of aperture of the lens is 5 cm ?

In a single slit diffraction pattern obtained on a screen, if the slit width is gradually decreased, which of the following statement is/are correct:A. Fringe pattern will expandB. Fringe pattern will shrinkC. Fringes may disappearD. No effect on fringe pattern

A plane wave of wavelength `6250 Å` is incident normally of the principal maximum on a screen distant 50 cm will beA. `312.5 xx 10^(-2)cm`B. `312.5xx10^(-4)cm`C. `312cm`D. `312.5xx10^(-5)cm`

The diameter of an objective of a telescope, which can just resolve two stars situated at angular displacement of `10^(-4)` degee, should be `(lambda = 5000 Å)`A. `35mm`B. `35cm`C. `35m`D. `3.5cm`

Angular width `(beta)` of central maximum of a diffraction pattern on a single slit does not depend uponA. (a) Distance between slit and sourceB. (b) Wavelength of the slitC. (c) Width of the slitD. (d) Frequency of light used

In Fresnel diffraction, if the distance between the disc and the screen is decreased, the intensity of central bright spot willA. (a) IncreaseB. (b) DecreaseC. (c) Remain constantD. (d) None of these

The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 Å. Then the slit width will beA. `5mm`B. `2.5mm`C. `1.25mm`D. `1.0mm`

The two slits are 1 mm apart from each other and illuminated with a light of wavelength `5xx10^(-7)` m. If the distance of the screen is 1 m from the slits, then the distance between third dark fringe and fifth bright fringe isA. 1.2 mmB. 0.75 mmC. 1.25 mmD. 0.625 mm

A single slit of width `0.20mm` is illuminated with light of wavelength `500nm`. The observing screen is placed `80cm` from the slit. The width of the central bright fringe will beA. (a) `1mm`B. (b) `2mm`C. (c) `4mm`D. (d) `5mm`

A parallel beam of light of wavelength `6000Å` gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :A. `6 xx 10^(-3)` radB. `1.8 xx 10^(-3)` radC. `3 xx 10^(-3)` radD. `2 xx 10^(-3)` rad

In single slit fraunhoffer diffraction which type of wavefront is required-A. cylindricalB. sphericalC. ellipticalD. plane

A ray of light is incident on the left vertical face of the glass slab. If the incident light has an intensity I and on each reflection the intensity decreases by 90% and on each refraction the intensity decreases by 10%, find the ratio of the intensities of maximum to minimum in reflected pattern.

The wavefront of a light beam is given by the equation `x + 2y + 3x = c` (where c is arbitrary constant), then the angle made by the direction of light with the y-axis isA. `cos^(-1).(1)/(sqrt(14))`B. `sin^(-1).(2)/(sqrt(14))`C. `cos^(-1).(2)/(sqrt(14))`D. `sin^(-1).(3)/(sqrt(14))`

A beam of light travelling in water falls on a glass plate immersed in water. When the incident angle is `51^(@)`, the reflected beam of light is found to be completely plane polarised. Determine the refractive index of glass. Given refractive idex of water =4/3.

When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light isA. (a) polarisedB. (b) of longer wavelengthC. (c) of shorter wavelengthD. (d) of high intensity

In YDSE, `d=5lambda`, then the total no. of maxima observed upon screen will beA. 9B. 8C. 7D. 5

A material having an index of refraction of 1.30 is used as an antireflective coating on a piece of glass `(n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 500 nm light?

If yellow light of `5890 Å` is incident on a grating that has 850 lines per mm. what will be the angle between both the first order images?

With a thin air film between two points, 6 fringes appear when light of wavelength 5890 Å is used. Calculate the difference in the thickness of the film between the two points.