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Quiz 12
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Q.1
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.2
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
0%
b) 45째
0%
c) 60째
0%
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.3
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
0%
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.4
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.5
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.6
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
0%
b) 1.30
0%
c) 1.36
0%
d) 1.42
Explanation
From figure n1=1.36 Answer:(c)
Q.7
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
0%
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.8
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
0%
b) 70 cm
0%
c) 80 cm
0%
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.9
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
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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.10
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.11
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
0%
c) l is dependent on n(z)
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b) n₁ sinθi = n₂ sinθf
0%
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.12
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.13
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]
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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.14
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
0%
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.15
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)
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b) For this prism, the refractive index μ and the angle of prism A are related
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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
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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.16
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
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b) The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant
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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)
Q.17
Light of λ=589 nm traverses a tank of height 20.06 m first filled with glycerine ( µ=1.47) and then carbon disulphide. The difference in the time taken to traverse the tank is 1.07× 10⁻⁸s. The refractive index of carbondisulphide is
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a) 1.6
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b) 1.62
0%
c)1.58
0%
d)1.63
Explanation
Answer: (d)
Q.18
Two slits separated by a distance of 1mm are illuminated with red light of wave length 6.5×10⁻⁷ m. The interference fringes are observed on screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to [ MPPMT 1995] [ MPPMT 1995]
0%
a) 0.65 mm
0%
b) 1.63 mm
0%
c) 3.25 mm
0%
d) 4.88 mm
Explanation
Position of third dark fringe Position of fifth bright fringe ∴ Difference in their positions=3.25 - 1.625=1.625=1.63 mmAnswer: (b)
Q.19
A slit of width 12×10⁻⁷ m is illuminated by light of wavelength 6000Å. The angular width of the central maximum is approximately
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a) 30°
0%
b) 60°
0%
c)90°
0%
d)0°
Explanation
Answer:(b)
Q.20
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
0%
c) 0.10 cm
0%
d) 0.25 cm
Explanation
Lens form image on screen placed at focal length f = D = 60 cm For first minima, Answer:(b)
Q.21
The diameter of the object lens of telescope is 5.0 m and wavelength of light is 6000 Å . The limit of resolution of this telescope is [AFMC 1997]
0%
a) 0.15 sec
0%
b) 0.06 sec
0%
c)0.03sec
0%
d)3.03 sec
Explanation
Angular limit of resolution=1.22λ / D Angular limit of resolution=1.22×6000×10⁻¹⁰ / 5 Angular limit of resolution=1464 ×10⁻¹⁰ rad1 rad=(360)(600)60)/2π sec=2.06× 105∴ Angular limit of resolution=0.03 sec Answer:(c)
Q.22
Light from a denser medium I goes to rarer medium II. When angle of incidence is θ, the reflected and refracted rays are perpendicular to each other. The critical angle is [ AFMC 2000]
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a)sin⁻¹(cos θ)
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b) sin⁻¹(cot θ)
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c)sin⁻¹(tan θ)
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d)sin⁻¹(1)
Explanation
light is travelling from denser medium refractive index 'n12’ then sinθ/ sin r= n2/n1 From figure r=90-θ sinθ/sin(90-θ)= n2/n1 sinθ/cosθ= n2/n1 ∴ tanθ= n2/n1 For critical angle sinC= n2/n1 tanθ=sinC C = sin⁻¹tanθ Answer: (c)
Q.23
The frequency of light wave in material is 2×1014Hz and wavelength is 5000Å. The refractive index of material will be... [ CBSE-PMT 2007]
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a) 1.50
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b) 3.00
0%
c) 1.33
0%
d) 1.40
Explanation
According to equation velocity of light in medium v =fλ here f is the frequency f = 2×1014 Hz λ = 5000Å =5000×10⁻¹⁰ m v = 2×1014 ×5000×10⁻¹⁰ v = 108 m/s Refractive index of material µ = c/v = 3×108 / 108 = 3 Answer: (b)
Q.24
Time taken by sunlight to pass through a window of thickness 4mm whose refractive index is 3/2 is .. [ CBSE-PMT 1993]
0%
b) 2×108 sec
0%
d)2×1011 sec
0%
a) 2×10⁻⁴ sec
0%
c)2×10⁻¹¹
Explanation
Velocity of light in glass vg=c / µvg=3×108 / 1.5=2×108 m/s Now if t is times thent=thickness/ velocityt=4×10⁻³ / 2×108t=2×10⁻¹¹ sec Answer:(c)
Q.25
ray of light from denser medium strikes a rare medium at an angle of incidence i ( see figure). The reflected and refracted ray's make an angle of 90° with each other. The angles of reflection and refraction are r and r'. The critical angle is [ IIT 1983]
0%
a) sin⁻¹(tan r)
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b) sin⁻¹(tan i)
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c) sin⁻¹(tan r')
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d) tan⁻¹(sin i)
Explanation
for critical angle sin c=n1 / n2 --eq(1) here n1 is refractive index of rarer medium and n2 is refractive index of denser medium From snells law from equation (1) and above equation sin C=tan r C=sin⁻¹(tan r) Answer: (a)
Q.26
A beam of light of wave length 600nm from a distance source falls on a single slit 1mm wide and a resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on ether side of central bright fringe is [ IIT 1994]
0%
a) 1.2 cm
0%
b) 1.2 mm
0%
c)2.4 cm
0%
d)2.4 mm
Explanation
The distance between the first dark fringe on either side of the central maximum=width of the central maximum=2Dλ / d=(2×2×600×10⁻⁹) / 10-3=2.4× 10⁻³ m=2.4 mm Answer:(d)
Q.27
An isosceles prism of angle 120° has a refractive index 1.Two parallel monochromatic rays enter the prism parallel to each other as shown. The rays emerges from the opposite faces [ IIT 1995]
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a) are parallel to each other
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b) are diverging
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c) makes an angle 2[sin-190.72) - 30°] with each other
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d) make an angle of 2sin-1(0.72) with each other
Explanation
Applying Snell's law at P sinr / sin30=n sin r=1.44 /2=0.72 ∴ δ=r -30 ∴ δ=sin⁻¹ (0.72) - 30° ∴ The rays makes an angle of 2δ=2[sin-190.72) - 30°] with each other Answer: (d)
Q.28
A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 ( n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incident αmax such that the ray comes out only from the other surface CD is given by [ IIT 2000]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
See figure. the ray will come out from CD if it suffers total internal reflection at surface AD. i.e. it strikes the surface AD at critical angle C Applying Snell's law at P n1sinC=n2 sinC=n2 / n1 C=sin⁻¹(n2 / n1)Applying Snell's law at Q n2α=n1cosCThus Answer: (a)
Q.29
In Yong's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is [ IIT 2005]
0%
a) sin⁻¹ (λ/d)
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b) sin⁻¹ (λ/2d)
0%
c)sin⁻¹ (λ/3d)
0%
d)sin⁻¹ (λ/4d)
Explanation
Let P be the point on the central maxima whose intensity is one fourth of the maximum intensity For interference we know that Now I1=I2=I and Imax=4I Thus I=I+I+2IcosΦ ⇒ -(1/2)=cosΦ ∴ Φ=2π/3 For a phase difference of 2π/ 3, the path difference is But the path difference ( in terms of P and Q ) is dsin θ as shown in figure ∴ d sinθ=λ/3∴ θ=sin⁻¹( λ / 3d)Answer: (c)
Q.30
A light beam is traveling from region I to IV( figure). The refractive index in regions I, II, II and IV are no , no/2, no/6, no/8 respectively. The angle of incident θ for which the beam just misses entering region IV is .. [ IIT 2008]
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a) sin⁻¹(3/4)
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b) sin⁻¹(1/8)
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c)sin⁻¹(1/4)
0%
d)sin⁻¹(1/3)
Explanation
Light ray do not enter region IV. It implies that angle of refraction must be 90° at the surface separating region III and IV Thus no sinθ=(no/8) sin 90sinθ=(1/8) θ=sin⁻¹(1/8) Answer:(b)
Q.31
The angle of incidence at which reflected light is totally polarized for reflection from air to glass ( refractive index n) is [ AIEEE 2004]
0%
a) tan⁻¹ (1/n)
0%
b) sin⁻¹(1/n)
0%
c) sin⁻¹(n)
0%
d) tan⁻¹ (n)
Explanation
The angle of incidence for total polarization is given by tanθ=n θ=tan⁻¹n Answer:(d)
Q.32
Two point white dots are 1 mm apart on blank paper. They are viewed by eye pupil diameter 3 mm. Approximately what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light=500 nm] [AIEEE 2005]
0%
a)1 m
0%
b)5 m
0%
c)3 m
0%
d)6 m
Explanation
Distance between two points y=1 mm=10-3 m Diameter of pupil d=3 mm=3×10⁻³ mwavelength of light λ=500 nm=5×10⁻⁷ From formula Answer:(b)
Q.33
If the refractive index of water is 4/3 and that of a given slab of glass immersed in it is 5/3, what will be the critical angle of incidence for a ray of light tending to go from glass to water [ Raj.PMT 1996]
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d)It can not be determined from the given data
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a) sin⁻¹(5/4)
0%
b) sin⁻¹(3/5)
0%
c)sin⁻¹(4/5)
Explanation
sinC=nw / ng sinC=(4/3) / (5/3)=4/5C=sin⁻¹ (4/5)Answer: (c)
Q.34
The average distance between the earth and moon is 38.6×104km. The minimum separation between the two points on the surface of moon that can be resolved by a telescope whose objective lens has a diameter of 5 metres with λ=6000 Angstrom, is [ MPPMT 1993]
0%
a)5.65 m
0%
b) 28.25 m
0%
c) 11.30 m
0%
d) 46.51 m
Explanation
Resolving limit of a telescope=λ/d=6×10⁻⁷ / 5 --eq(1)Resolving limit=separation / average distance between moon and earth --eq(2) 6×10⁻⁷ / 5=x/ 38.6×107 x=38.6×107×1.2×10⁻⁷=46.32 mAnswer: (d)
Q.35
Light of wavelength 6328Å is incident on slit having a width of 0.2mm. the width of the central maximum, measured from minimum to minimum of diffraction pattern on a screen 9.0 m away will be about [ MPPMT 1987]
0%
a) 0.36°
0%
b) 0.18°
0%
c)0.72°
0%
d)0.09°
Explanation
Angular width of central maxima θ=2λ /d=θ=2×6328×10⁻¹⁰ / 0.2×10⁻³ θ=6328×10⁻⁶ θ=0.36°Answer: (a)
Q.36
A beam of light of wavelength 600nm from a distance source falls on a single slit 1.00mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is [ IIT 1994]
0%
a)1.2cm
0%
b) 1.2 mm
0%
c)2.4 cm
0%
d)2.4 mm
Explanation
θ=2λ/d and θ=x/D X=D×2λ / dX=2×2×600×10⁻⁹ / 10-3 x=2400×10⁻⁶ m x=2.4mmAnswer: (d)
Q.37
A single slit is located effectively at infinity in front of lens of focal length 1m, and it is illuminated normally with light of wavelength 600nm. The first minimum on either side of central maximum are separated by 4mm. Width of the slit is [ Kerala PMT 2004]
0%
a) 0.1 mm
0%
b) 0.2 mm
0%
c) 0.3 mm
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d) 0.4 mm
Explanation
Here f=D=1 m, given 2w=4 mm , x=2mm width of fringe x=Dλ /d d=Dλ /x d=1×6×107 / 2×10⁻³ d=3×10⁻⁴=0.3mm Answer: (c)
Q.38
A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between first dark fringes on either side of the central bright fringe is
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a) 2.4 cm
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b) 2.4mm
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c) 1.2 cm
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d) 1.2mm
Explanation
Distance between first dark fringes = width of central bright fringe (2x) 2x=2.400×10⁻³ m = 2.4 mm Answer:(b)
Q.39
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]
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a)
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b)
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c)
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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.40
A point object O is placed at a distance of 20cm from a convex lens of focal length 10cm as shown in figure. At what distance x from lens should a convex mirror of focal length 60 cm, be laced so that final image coincides with object?
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a) 10 cm
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b) 40 cm
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c)20 cm
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d)final image can never coincide with the object under the given condition
Explanation
Object is placed at distance 2f from the lens . the image is formed at 2f or 2 cm from lens. So, if the convex mirror is placed at 20 cm from lens , the first image formed by lens will be at the pole of mirror and it will reflect all the rays symmetrically to other side as shown in figure and the final image will coincide with the object Answer:(c)
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