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Quiz 12
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Q.1
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.The focal length of the combination is : … [ AIMPT 2015]
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a) 50 cm
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b) -20 cm
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c) -25 cm
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d) -50 cm
Explanation
Liquid in between two lens acts as concave lens. Combination of three lenses. Answer:(a)
Q.2
For a parallel beam of monochromatic light of wavelength ‘λ‘, diffraction is produced by a single slit whose width ‘a‘ is of the order of the wavelength of the light. If ‘D‘ is the distance of the screen from the slit, the width of the central maxima will be :
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a)
0%
b)
0%
c)
0%
d)
Explanation
Standard formula for width of central maxima Answer:(b)
Q.3
At the first minimum adjacent to the central maximum of a single slit diffraction pattern the phase difference between the Huygen鈥檚 wavelet from the edge of the slit and the wavelet from the mid point 鈥 Re AIPMT 2015]
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a) π/8 radian
0%
b) π/4 radian
0%
c) π/2 radian
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d) π radian
Explanation
From the figure path difference is 位/2 thus phase difference is 蟺 radian Answer:(d)
Q.4
Two slits in Youngs experiment have widths in the ratio 1 :The ratio of intensity at the maxima and minima in the interference pattern, Imax/Iminis :
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a) 4/9
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b) 9/4
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c) 121/49
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d) 49/121
Explanation
Answer:(b)
Q.5
In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is :…[ Re AIPMT 2015]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Since real image is formed ray diagram will be Magnification (M) is ratio of image height and object height M=I/L Answer:(a)
Q.6
A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39,1.44 and 1.47, respectively. The prism will :- [ ReAIPMT 2015]
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a) separate the red colour part from the green and blue colours
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b) separate the blue colour part from the red and green colours
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c) separate all the three colours from one another
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d) not separate the three colours at all
Explanation
Angel of incidence is 45° n=1.41 more the refractive index, lesser is the critical angle. That is if angle of incidence is more than critical angle light will internally reflected Thus for any light having refractive index more than 1.41 will be totally internal reflected because critical angle is less than 45 Red light refractive index < 1.41. Thus red will pass others will totally internally reflected Answer:(a)
Q.7
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of…[ AIPMT 2016]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Condition for minima: general equation is asinθ = nλ First minimum dsinθ=λ asin30=5000×10-10 a= 10-6 m General maxima n=1 First secondary maxima Answer:(d)
Q.8
The intensity at the maximum in a Young's double slit experiment is IDistance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?[ AIPMT 2016]
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a) I0
0%
b) I0/4
0%
c) 3I0/4
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d) I0/2
Explanation
It is given that that maximum intensity which is 4I0 in above equation should now be treated as I0 equation changes to K=2×/λ ,δ = path difference Here x= 2.5λ and d =5λ, D= 50λ I = I0/2 Answer:(d)
Q.9
The angle of incidence for a ray of light at a refracting surface of a prism is 45° . The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are : … [ AIPMR 2016]
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a) 45°,1/√2
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b) 30°, √2
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c) 45°, √2
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d) 30°,1/√2
Explanation
i = 45° ; A = 60°; δm = 2i – A = 30° Answer:(b)
Q.10
A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :-
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a) 37.3 cm
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b) 46.0 cm
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c) 50.0 cm
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d) 54.0 cm
Explanation
Using lens formula for objective lens v0= 50 cm Tube length l= |v0| + fe = 50 +4 = 54 cm Answer:(d)
Q.11
Two identical glass (μg=3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μw=4/3). The focal length of the combination is ….[NEET II -2016]
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a) 4f/3
0%
b) 3f/4
0%
c) f/3
0%
d) f
Explanation
Power of lenses gets added Given that both convex lenses have focal length =f Liquid between the convex lenses form concave lens Answer:(b)
Q.12
An air bubble in a glass slab with refractive index1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is …[ NEET II- 2016]
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a) 12
0%
b) 16
0%
c) 8
0%
d) 10
Explanation
ho = real depth, hi = apparent depth no= Refractive index of medium in which object is ni = Refractive index of medium in which observer is First side ho = 7.5 cm Second side h’o = 4.5 cm Thickness of slab =7.5 +4.5 = 12cm Answer:(a)
Q.13
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio will be…[ NEET II -2016]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
nI1=I2 Answer:(d)
Q.14
A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be :-
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a) concave, – 0.2 diopter
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b) convex, + 0.15 diopter
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c) convex, + 2.25 diopter
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d) concave, – 0.25 diopter
Explanation
As we want to correct myopia. So, far point must go to infinity. v = –4 m, u = – ∞, P = ? (–) implies concave mirror Answer:(d)
Q.15
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-
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a) 0.20 cm
0%
b) 0.15 cm
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c) 0.10 cm
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d) 0.25 cm
Explanation
Lens form image on screen placed at focal length f = D = 60 cm For first minima, Answer:(b)
Q.16
Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5thdark fringe lies in air. The refractive index of the medium is nearly
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a) 1.25
0%
b) 1.59
0%
c) 1.69
0%
d) 1.78
Explanation
In liquid wavelength decreases given by relation nλ’ =λ …(i) Position is same or x is same for both fringes, D and d same 5thdark fringe 8th bright fringe From (i) n = 16/9 = 1.777 = 1.78 Answer:(d)
Q.17
The ratio of resolving powers of an optical microscope for two wavelengths λ1= 4000 Å and λ2= 6000 Å is …[NEET 2017]
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a) 8 : 27
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b) 9 : 4
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c) 3 : 2
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d) 16 : 81
Explanation
Answer:(c)
Q.18
Two Polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarised light I0 is incident on PA third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45° with that of PThe intensity of transmitted light through P2 is …
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a) I0/2
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b) I0/4
0%
c) I0/8
0%
d) I0/16
Explanation
Intensity of light passing through P1 = I0/2 Now P3 is at angle 45° with P3 thus I’ Now P4 is at angle of 45° with P3 thus I” Answer:(c)
Q.19
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by … [ NEET 2017]
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a) y/2x
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b) x/y
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c) x/2y
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d) x/y
Explanation
When the mirror is rotated by θ then angle between the incident and reflected ray 2θ. Given that angle θθ is very small, 2θ is also small, M is mirror , N is normal to mirror From geometry y = x 2θ θ=y/2x Answer:(a)
Q.20
A thin prism having refracting angle 10° is made of glass of refractive index 1.This prism is combined with another thin prism of glass of refractive index1.This combination produces dispersion without deviation. The refracting angle of second prism should be
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a) 4°
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b) 6°
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c) 8°
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d) 10°
Explanation
For small angle prism minimum deviation is given by δm = (n1 – 1) A Since combination of two prism do not produce deviation (n1 – 1) A +(n2 – 1) A’ = 0 |(1.42-1) 10| = |(1.7 – 1) A’| 4.2 =0.7 A’ Answer:(b)
Q.21
A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is :
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a) I0
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b) I0/2
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c) I0/4
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d) I0/8
Explanation
Intensity of light is halved upon passage through first polaroid. Thus intensity of light incident of B I’= I0/2 which is at 45° with A Using Malus’ Law : Answer:(c)
Q.22
Two coherent point sources S1 and S2 are separated by a small distance 'd' as shown. The fringes obtained on the screen will be …[ IIT Mains 2013]
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c) semi−circles
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a) points
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b) straight lines
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d) concentric circles
Explanation
Wave-front emitted will be spherical thus concentric circles Answer:(d)
Q.23
The graph between angle of deviation δ and angle of incidence I for triangular prism is represented by
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer:(c)
Q.24
In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is … [ IIT Advance 2013]
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a)
0%
b)
0%
c)
0%
d)
Explanation
Possible if Answer:(b)
Q.25
The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is …. [ IIT Advance 2013]
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a) 1 m
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b) 2m
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c) 3m
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d) 6m
Explanation
Refractive index of glass Magnification m=v/u Given: real image thus m = -1/3 and image distance v = 8 cm ∴ u = -24 cm From lens maker formula f = 6 cm R = 3cm Answer:(c)
Q.26
A ray of light travelling in the direction is incident on a plane mirror. After reflection, it travels along the direction . The angle of incidence is ..[ [IIT Advanced 2013]
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a) 30째
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b) 45째
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c) 60째
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d) 75째
Explanation
Angle between two rays ∴ θ = 120° is angle between incident and reflected ray Now angle of incidence = angle of reflection Thus angle of incidence = 60° Answer:(c)
Q.27
A right angled prism of refractive index µ1 is placed in a rectangular block of refractive index µ2, which is surrounded by a medium of refractive index µ3, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between µ1, µ2 and µ3, it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'. Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists: [IIT Advance 2013]
List I
List II
P. e → f
μ
1
> 2 μ
2
Q. e → g
µ
2
> µ
1
and µ
2
> µ
3
R. e → h
µ
1
= µ
2
S. e → i
µ
2
< µ
1
< 2µ
2
and µ
2
> µ
3
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a) P → 2 ; Q → 3; R → 1; S → 4
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b) P → 1 ; Q → 2; R → 4; S → 3
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c) P → 4 ; Q → 1; R → 2; S → 3
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d) P → 2 ; Q → 3; R → 4; S → 1
Explanation
(P) e → f → for 1 → 2 towards normal ⇒ µ2 > µ1 For 2 → 3 always from normal ⇒ µ3 < µ2 P → 2 (Q) e → g → no deviation any where µ1 = µ2 = µ3 Q → 3 (R) e → h → 1 → 2 away from normal.µ2 < µ1 2 → 3 away from normal.⇒ µ3 < µ2 And also at 1 − 2 no I.R. ⇒ µ1 < √2 µ2 ⇒ µ2 < µ1 < 2µ2 Q → 4 (S) e → i → total internal reflection at 1 → 2 S → 1 (P) → 2; (Q) → 3; (R) → (4); (S) → (1). Answer:(d)
Q.28
than one correct answer Q578) A light source, which emits two wavelengths λ1 = 400 nm and λ2 = 600 nm, is used in a Young's double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum arem1 and m2, respectively, then ..[ IIT Advance 2014]
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a) β2 > β1
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b) m1 > m2
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c) From the central maximum, 3rd maximum of λ2 overlaps with 5th minimum of λ1
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d) The angular separation of fringes for λ1 is greater than λ2
Explanation
Fringe width β= λD/d β1 < β2 Option “a” correct Since fringe width of λ1 is less than λ2 number of fringes of λ1 (m1) will be more than m2 . Option “b” is correct 5th minimum of λ1 3rd maximum of λ2 If overlap then Now It is given λ2 = 600 nm thus option c is true Option d Since β1 < β2 angular width or angular separation of fringes for λ1 is smaller than λ2 Option d is wrong Answer:(a, b, c)
Q.29
A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then … [ IIT Advance 2014]
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a) | f1 | = 3R
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b) | f1 | = 2.8 R
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c) | f2 | = 2R
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d) | f2 | = 1.4R
Explanation
For film u= ∞ It will act as object for glass f1 = 3R ( option a correct) while rays of light traversing from glass to air for glass u=∞ V=14R For film u = 14R f2 = 2R option “c” correct Answer:(a, c)
Q.30
A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is …[ IIT Advance 2014]
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a) 1.21
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b) 1.30
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c) 1.36
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d) 1.42
Explanation
From figure n1=1.36 Answer:(c)
Q.31
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.
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a) P-1, Q-2, R-3, S-4
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b) P-2, Q-4, R-3, S-1
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c) P-4, Q-1, R-2, S-3
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d) P-2, Q-1, R-3, S-4
Explanation
List I P F = r/2 P → 2 List I Q F=r; Q→ 4 List I R F = -r ;R→3 List S F= 2r; S → 1 P → 2 ; Q → 4; R →3; S →1 Answer:(b)
Q.32
Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside SThe distance d is ,,[IIT Advance 2015]
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a) 60 cm
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b) 70 cm
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c) 80 cm
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d) 90 cm
Explanation
Image formed by S1 in air v = 50cm it acts as object for S1 Since rays are parallel to principle axis in S2 image is at ∞ d -50= 20 d=70 Answer:(b)
Q.33
Paragraph Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index nThe light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index nThe numerical aperture (NA) of the structure is defined as sinim. [ IIT Advance 2015] Q587A)More than one correct answer For two structures namely S1 with and S2 with n1=8/5 and n2=7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)
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a) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16/(3√15)
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b) NA of S1 immersed in liquid of refractive index 6/√15 is the same as that of S2 immersed in water.
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c) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 4/√15
0%
d) NA of S1 placed in air is the same as that of S2 placed in water.
Explanation
Given NA2 < NA1 Now angle of emergence from NA1 will have angle i1 which will be more than maximum angle of incidence i2 for NA2 Thus combined structure will have numerical aperture of NA2 Correct option d Answer:(d)
Q.34
Right angled triangular prism of refractive index n =√2 Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45°. The angle θ of the prism is … [ IIT Advance 2016]
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a) 15°
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b) 22.5°
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c) 30°
0%
d) 45°
Explanation
∴ r1 = 30° Since light totally internally reflected thus Sinr2 = 1/√2 ∴ r2=45° From the triangle PAM θ = r2 –r1 = 45-30= 15° Answer:(a)
Q.35
A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2(> n1), as shown in the figure. A ray of light is incident with angle θ1 from medium 1 and emerges in medium 2 with refraction and θf with a lateral displacement l …[IIT Advance 2016]] Which of the following statement(s) is(are) true?
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a) l is independent of n2
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b) n1 sinθi = n2 sinθf
0%
c) l is dependent on n(z)
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d) n1 sin θi = (n2 - n1) sinθ
Explanation
Deviation l depends on n(z) hence is independent of n2 option a and c is correct According to Snel’ls law option b is correct and option d is wrong Answer:(a,b,c)
Q.36
than one correct option Q590) A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?
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a) The refractive index of the lens is 2.5
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b) The radius of curvature of the convex surface is 45 cm
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c) The faint image is erect and real
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d) The focal length of the lens is 20 cm
Explanation
As size of image is two times u=-30, v = 60 From lens formula Hence f = 20 cm Option d correct Image form by reflection at 10 cm away because of convex surface hence image is virtual and erect option “c” is wrong u= - 30 v = 10cm From mirror formula For mirror R = 2f = 30 cm option b wrong Now from formula n= 2.5 option a correct Answer:(a,d)
Q.37
A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ = 30θ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are [IIT Advance 2016]
0%
a) (25, 25√3)
0%
b) (0, 0)
0%
c) (125/3, 25/√3 )
0%
d) (50 -25√3, 25)
Explanation
Image formed at v V = 75 Distance of point on principle axis is at 75-50 = 25 cm Image form by convex lens V= 50 cm ( x=50 cm and y = 0 ) For sake of simplicity we will consider plane mirror tilted by 30° Then image will deviate by 2θ = 60° Thus coordinates will be x-xcosθ = 50 –50cos60 = 50-25 =25 Y coordinate = xsin60 =50sin60 = 25√3 Answer:(a)
Q.38
than one correct answer Q592) While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1SWhich of the following is(are) true of the intensity pattern on the screen? [ IIT Advance 2016]
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a) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
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b) Straight bright and dark bands parallel to the x-axis
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c) Semi-circular bright and dark bands centered at point O
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d) The region very close to the point O will be dark
Explanation
Option a At ‘O’ Path difference = 0.6003 = nλ n is not integer thus not a constructive interference at “O” Path difference = 0.6003 = (2n-1)λ/2 n is nteger thus destructive interference at “O” option a wrong, option c is correct Option B For a point on screen at P(x,z) have certain path difference. Will remain unchanged fro all points which follows r2 = x2 + z2. Thus semicircular band, half circular bands will be below O Answer:(c, d)
Q.39
For an isosceles prism of angle A and refractive index µ it is found that the angle of minimum deviation δm = A. Which of the following options is/are correct ? [ IIT Advance 2016]
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a) a) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by r1 = (i1/2)
0%
b) For this prism, the refractive index μ and the angle of prism A are related
0%
c) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is
0%
d) For the angle of incidence i1 = A, the ray inside the prism is parallel to the base of the prism.
Explanation
Given δm = A., at angle of minimum deviation r1=r2 Thus 2r1 = A, i= e and δm = A. = 2i-A Given δm = A. there fore A= 2i-A or i= A [Option d correct As 2r1 = A therefore r1 = i/2 [ Option a correct,] [Option b wrong] Option c Angle of emergence = 90° µSinr2 = 1×sin90 µSinr2 = 1 Now sini= μsinr1 As r1 + r2 = A From (ii) Option c is correct Option d When i=e then ray is parallel to base Answer:(a, b, c)
Q.40
Two coherent monochromatic point sources S1 and S2 of wavelength λ= 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ. Which of the following options is/are correct ?
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a) A dark spot will be formed at the point P2
0%
b) The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant
0%
c) At P2 the order of the fringe will be maximum
0%
d) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000
Explanation
For bright spot at P1 ; S1P1 - S2P2= nλ d= nλ 1.8=600×10(-6) n ∴ n = 3 × 103 n Is integer thus a is wrong For bright spot at P2 n = 3000 [Option c correct] Thus total fringes in first quadrant are 3000 [ option d is correct] Path difference PD=(2n-1) λ/2 Path difference Thus angular width increases as we move from P1 to P2 along the first quadrant Option b wrong Answer:(c, d)
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