1 Crore+ students have signed up on EduRev. Have you? 
An opaque sphere of radius a is just immersed in a transparent liquid as shown in figure. A point source is placed on the vertical diameter of the sphere at a distance a/2 from the top of the sphere. One ray originating from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is 30º. The refractive index of the liquid is
A thin oil film of refractive index 1.2 floats on the surface of water (m = 4/3) When a light of wavelength l = 9.6 × 10^{–7} m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is:
In a YDSE experiment if a slab whose refractive index can be varied is placed in front of one of the slits then the variation of resultant intensity at midpoint of screen with 'µ' will be best represented by (µ ³ 1).[ Assume slits of equal width and there is no absorption by slab; mid point of screen is the point where waves interfere with zero phase difference in absence of slab]
In absence of film or for μ=0 intensity is maximum at screen. As the value of μ is increased, intensity shall decrease and then
increase alternately. Hence the correct variation is
Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made reflecting. The acceleration of two images
formed in those two reflecting surfaces w.r.t. each other is:
A ray is incident on the first prism at an angle of incidence 53º as shown in the figure. The angle between side CA and B'A' for the net deviation by both the prisms to be double of the deviation produced by the first prism, will be :
Let be the angle of emergence from the first prism be ‘e’ Snell's law on surface AB
Then for net deviation to be double, the incident ray on side
A' B' of second prism should make angles i or e with normal.
Hence the angle between the given then will be 2e or i + e.
Light of wavelength 4000 Å is incident at small angle on a prism of apex angle 4º. The prism has n_{v} = 1.5 & n_{r} = 1.48. The angle of dispersion produced by the prism in this light is :
Two plane mirrors are joined together as shown in figure. Two point objects O_{1} and O_{2} are placed symmetrically such that AO_{1} = AO_{2}. The image of the two objects is common if :
Monochromatic light rays parallel to xaxis strike a convex lens AB. If the lens oscillates such that AB tilts upto a small angle q (in radian) on either side of yaxis, then the amplitude of oscillation of image will be (f = focal length of the lens) :
From the figure :
A particle is moving towards a fixed convex mirror. The image also moves. If V_{i} = speed of image and V_{O} = speed of the object, then
The speed of light in the material of a plano convex lens is 2 × 108 m/s and its greatest thickness is 3 mm. If the aperture diameter of the lens in 6.0 cm then :
In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images formed by a convex lens placed between them is 4.84.
Which of the following statements is/are correct about the refraction of light from a plane surface when light ray is incident in denser medium. [C is critical angle]
Statement1 : A beam of white light enters the curved surface of a semicircular piece of glass along the normal. The incoming beam is moved clockwise (so that the angle q increases), such that the beam always enters along the normal to the curved side. Just before the refracted beam disappears, it becomes predominantly red.
Statement2 : The index of refraction for light at the red end of the visible spectrum is more than at the violet end
by cauchy's formula
: a & b are constant of medium. The index of refraction for light at the red end
of the visible spectrum is lesser than at the violet end.Hence statement 2 is false
Statement1 : A point object moves near the principal axis of a fixed spherical mirror along a straight line. Then the image formed by the spherical mirror also moves along a straight line.
Statement2 : For an incident ray on a fixed spherical mirror there is a fixed reflected ray. If a point object moves along this incident ray, its image will always lie on the given reflected ray. Further an incident ray may be drawn from the moving point object in its direction of velocity towards the mirror.
Draw an incident ray on the mirror and trace the corresponding reflected ray. If a point object moves along this ray,its image will always lies on the traced reflected ray. Hence when a point object moves near the principal axisof a fixed spherical mirror along a straight line, then its image formed by the spherical mirror also moves along
a straight line.Statement1 is True, Statement2 is True; Statement2 is a correct explanation for Statement1.
Paragraph for Question Nos. 15
A monochromatic point source S of wavelength l = 5000√2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along yaxis at horizontal distance D = 1 metre from point source. The refractive index of liquid is √2 . ( Neglect partial reflection of rays from source S at liquidair interface.)
Q. The ycoordinates of second order bright formed on the screen is
The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflected back (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
So Interference pattern is formed in region OP.Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0
Paragraph for Question Nos. 15
A monochromatic point source S of wavelength l = 5000 √2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along yaxis at horizontal distance D = 1 metre from point source. The refractive index of liquid is√2 . ( Neglect partial reflection of rays from source S at liquidair interface.)
Q. The region on screen where interference pattern is formed lies in
The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflectedback (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
Interference pattern is formed in region OP.
Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0.
Paragraph for Question Nos. 15
A monochromatic point source S of wavelength l = 5000√2 Å (in air) is placed at a distance d = 1 mm below the surface of transparent liquid as shown in figure. A very large screen is placed along yaxis at horizontal distance D = 1 metre from point source. The refractive index of liquid is√2 . ( Neglect partial reflection of rays from source S at liquidair interface.)
Q. The minimum horizontal distance of screen from fixed source such that at least one bright is formed on screen
The critical angle is C = 45°
Hence all rays from S which are incident on interface at an angle of incidence greater than 45° are reflected back (and appear to come from S1) and fall on screen between OP. Light also directly falls on OP from S.
Interference pattern is formed in region OP.
Since waves from S do not under go phase change during reflection, We can take S and S1 to be coherent sources in same phase. Hence zero order bright is formed at y = 0.
Matrix  Match Type
A white light ray is incident on a glass prism, and it create four refracted rays A, B, C and D. Match the refracted rays with the colours given (1 & D are rays due to total internal reflection.):
Ray Colour
(A) A (p) red
(B) B (q) green
(C) C (r) yellow
(D) D (s) blue
By snell's law n = and n = So at inclined face angular incidence is same for all ray and refraction is from denser to rarer. So emergent angle will be less for red.
An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m_{1} and m_{2} and placed at the end of the tube. m_{1} & m_{2} are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 10^{14} Hz. The light reflected from m_{1} and m_{2} forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m_{1} & m_{2}.
The position of the image formed by lens water combination is a × 10^{2}m from lens and the distance between the images formed by m_{1} & m_{2} is b × 10^{3} m then find ab/40.
.
In the figure shown light of wave length l = 5000 Å is incident on the slits (in a horizontal fixed plane) S_{1} and
S_{2} separated by distance d = 1 mm. A horizontal screen ‘S’ is released from rest from initial distance
D_{0} = 1 m from the plane of the slits. Taking origin at O and positive x and y axis as shown, at t = 2 seconds;
(Use g = 10 m/s^{2}) velocity is 10n j ˆ and acceleration is m j ˆ of central maxima, find the value of [mm/25].
A is a thin walled sphere at rest made up of glass. The radius of the sphere is 1m and it is filled with a
transparent liquid of refractive index m. S is the luminous source moving directly towards a plane
vertical mirror M. A fish in the liquid is moving towards S. The eye of the fish and S are collinear &
perpendicular to mirror M. At the moment S is 3m away from the centre of the sphere, fish observes
that image of S due to reflection is moving with speed of 13m/s. If speed of the fish relative to the
sphere is 10m/s and m = 1.5 then find the speed of the source at that instant. The system is placed
in air.
A light ray parallel to the principal axis is incident (as shown in the figure) on a thin planoconvex lens
with radius of curvature of its curved part equal to 10 cm. Assuming that the refractive index of the
material of the lens is 4/3 and medium on both sides of the lens is air, find the distance of the point from
the lens where this ray meets the principal axis. Find your answer in the form X/7 cm and fill value
of X.
On vigorous oxidation of (CH_{3})_{2}C=CH—CH_{2} —CH_{3} by permanganate solution gives :
Which of the following reactions give alkylation product :
In (A) acid cannot give iodoform reaction
(B) lacks in αhydrogen. It give cannizzaro. (C) It can give haloform reaction because of CCl_{3} group.
Statement1 : When benzene diazonium chloride is heated with H_{2}O, PhOH is formed and rate of this ArSN reaction can be increased by adding +m group at para position of the .
Statement2 : + m group generally increases the stability of carbocation.
Due to resonance or +m group, rate of ArSN does not increases
Statement1 : Nitration of aniline can be conveniently done by protecting the amino group by acetylation.
Statement2 : Acetylation decrease the electron density in the benzene ring and thereby prevents oxidation of benzene ring.
(A) If there is no αhydrogen atom than aldehyde gives cannizaro reactions.
(B) & (C) If αhydrogen atoms are presents and the at least one –CH_{3} groups are presents then show aldol and
haloform reaction.
(D) By oxidation followed by haloform reaction.
Consider the following conversion In order to carry out this conversion, five distinct reagents numbered 1 to 5 are to be used in successive
steps I to V. Reagents are numbred as Indicate the reagents that need to be used in successive steps I to V by indicating their number in successive
five boxes as seen below.
(For eg. If you think that reagent 2 is required in step I, reagent 4 is requried in step II, reagent 5 in step III,
reagent 1 in step IV and reagent 3 in step V, then your answer will be
Consider product C, F, G & step VII and follow the given instructions to fill the box
Where 'x' 'y' 'z' and 'w' are numerals If C can undergo cannizzaro reaction, write 1 in place of x, if it can undergo reformatsky reaction write 2 in place of x, if it can undergo michael addition write 3 in place of x.
If 'F' cannot show stereoisomerism write 0 (zero) in place of y, if F can show geometrical isomerism write 1
in place of y, if F can show optical isomerism write 2 in place y If G cannot show stereoisomeris write 0 in place of Z, if G can show geometrical isomerism write 1 in place of Z, if G cannot show optical isomerism write 2 in place Z. If step VII is neither a regiospecific nor stereospecific reaction write 0 in place of w, if step VII is stereospecific reaction then write 1 & if step VII is regosepecific reaction then write 2 in place of w.
Let f be a differentiable function satisfying f(xy) = f(x) . f(y),
Let f'(x) is non decreasing function for which f'(0) = 0, f(0) = 5 and f''(x) – f'(x)
if then number of solutions of the equation (where {.} represents fractional part function)
clear from the graph
Number of solutions for x between 3 and 15 if where [.] represents greatest integer
function, is
Area bounded by y = f ^{1}(x), x = 0, y = & y = , where f(x) = x + sin x, is
Compute the area of the figure bounded by straight lines x = 0, x = 2 and the curves y = 2^{x} and y = 2x – x^{2} is
If f(x) = ae^{2x} + be^{x} + cx satisfies the conditions f(0) = – 1, f'(ln 2) = 31,
STATEMENT1 : The area bounded by the curve x + y = a (a > 0) is 2a^{2} and area bounded
px + qy + qx – py = a, where p^{2} + q^{2} = 1, is also 2a^{2}.
STATEMENT2 : Since αx + βy = 0 is perpendicular to βx – αy = 0, we can take one as xaxis and another
as yaxis and therefore the area bounded by αx + βy + βx – αy = a is 2a^{2} for all α, β€ R,
α 0, β 0.
STATEMENT1 :
Then the axis get rotated through an angle θ,
the equation of the given curve becomes U + V = a the area bounded = 2a^{2}.
so statement1 is true
STATEMENT2 : The equation of the curve is
STATEMENT 1 :
STATEMENT 2 :
(where [.] represents greater integer function)
Consider the function defined implicitly by the equation Where [x] denotes the greatest integer function
Q. The area of the region bounded by the curve & the line x = – 1 is
Consider the function defined implicitly by the equation . Where [x] denotes the greatest integer function
Q. Line x = 0 divides the region mention above in two parts. The ratio of area of left hand side of line to right hand side of line is
Consider the function defined implicitly by the equation Where [x] denotes the greatest integer function
Q.
Solution of the differential equation then find the value of λ.
The area of the figure bounded by a curve, the xaxis, and two ordinates, one of which is constant, the other
variable, is equal to the ratio of the cube of the variable ordinate to the variable abscissa and equation of curve
is (λy^{2} –x^{2})^{3} = Cx^{2} , then find the value of λ.
2 videos324 docs160 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
2 videos324 docs160 tests








