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# MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics

## 59 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics

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This mock test of MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics for JEE helps you for every JEE entrance exam. This contains 59 Multiple Choice Questions for JEE MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics (mcq) to study with solutions a complete question bank. The solved questions answers in this MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics quiz give you a good mix of easy questions and tough questions. JEE students definitely take this MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics exercise for a better result in the exam. You can find other MCQs Of Ray And Wave Optics, Past Year Questions JEE Main, Class 12, Physics extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The line ab and cd represent waveforms of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium2 after refraction. Q. Light travels as a

Solution:

For plane wave fronts the beam of light is parallel.

QUESTION: 2

### The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The line ab and cd represent waveforms of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium2 after refraction. Q. The phases of the light wave at c, d, e and f are φc, φd, φe and φf respectively. It is given that φc ≠ φf.

Solution:

Since points c and d are on the same wavefront , therefore φd = φc

QUESTION: 3

### The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The line ab and cd represent waveforms of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium2 after refraction. Q. Speed of light is

Solution:

The gap between consecutive wavefronts in medium 2 is less than that is medium 1. Therefore, wavelength of light in medium 2 is less than that in medium 1. Therefore, speed of light is more in medium 1 and less in medium 2.

QUESTION: 4

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell’s law,  it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n= c /  where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.
In normal materials, both εr and μr , are positive, implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light.
However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

Q. For light incident from air on a meta-material, the appropriate ray diagram is

Solution:

As n is negative, therefore direction charges

n1 sin θ1 = n2 sin θ2

QUESTION: 5

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell’s law,  it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n= c /  where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.
In normal materials, both εr and μr , are positive, implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light.
However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

Q. Choose the correct statement.

Solution:

The physical characteristics remain unchanged.

*Multiple options can be correct
QUESTION: 6

Light guidance in an optical fibre 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 n2.
The 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 n1. The numerical aperture (NA) of the structure is defined as sin im.

Q. For two structure namely S1 with  and n2 = 3/2, 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)

Solution:

Applying Snell’s law at P; ns sin im = n1 sin (90° – C) ns = Refractive index of surrounding

(a), (c) are correct options

QUESTION: 7

Light guidance in an optical fibre 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 n2.
The 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 n1. The numerical aperture (NA) of the structure is defined as sin im.

Q. If two structure of same cross-sectional area, but different numerical apertures NA1 and NA2(NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is

Solution:

Here NA2 < NA1
∴ the NA of combin ed structure is equal to the smaller value of the two numerical apertures.
(d) is the correct option.

QUESTION: 8

STATEMENT-1 The formula connecting u, v and f for a spherical mirror is valid for mirrors whose sizes are very small compared to their radii of curvature. because

STATEMENT-2 Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

Solution:

Statement 1 :

NOTE : The mirror (spherical) formula  is valid only for mirrors of small apertures where the size of aperture is very small as compared to the radius of curvature of the mirror. This statement is true.
Statement 2 :

NOTE : Laws of mirror are valid for plane as well as large spherical surfaces.
Therefore, statement 2 is wrong.

QUESTION: 9

An astronomical telescope has a large aperture to

Solution:

KEY CONCEPT :  The resolving power of a telescope

where D = diameter of the objective lens

λ = wavelength of light.

Clearly, larger the aperture, larger is the value of  D, more is the resolving power or resolution.

QUESTION: 10

If two mirrors are kept at 60° to each other, then the number of images formed by them is

Solution:

KEY CONCEPT :  When two plane mirrors are inclined at each other at an angle θ then the number of the images of a point object placed between the plane mirrors is

∴ Number of images formed

QUESTION: 11

Electromagnetic waves are transverse in nature is evident by

Solution:

The ph enomenon of polarisation is shown only by transverse waves.

QUESTION: 12

Wavelength of light used in an optical in strument are λ1= 4000 Å and λ2 = 5000 Å , then ratio of their respective resolving powers (corresponding to λ1 and λ2) is

Solution:

QUESTION: 13

Which of the following is used in optical fibres?

Solution:

In an optical fibre, light is sent through the fibre without any loss by the phenomenon of total internal reflection as shown in the figure.

QUESTION: 14

Consider telecommunication through optical fibres. Which of the following statements is not true?

Solution:

Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference.

QUESTION: 15

To demonstrate the phenomenon of interference, we require two sources which emit radiation

Solution:

For the phenomenon of interference we require two sources of light of same frequency and having a definite phase relationship (a phase relationship that does not change with time)

QUESTION: 16

The image formed by an objective of a compound microscope is

Solution:

A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.

QUESTION: 17

To get three images of a single object, one should have two plane mirrors at an angle of

Solution:

is an even number. The number of images formed is given by

QUESTION: 18

A light ray is incident perpendicularly to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n

Solution:

The incident angle is 45° .
Incident angle > critical angle, i > ic

QUESTION: 19

A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object

Solution:

KEY CONCEPT :  The focal length(F) of the final mirror

The combination acts as a converging mirror. For the object to be of the same size of mirror, u = 2F = 20 cm

QUESTION: 20

The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index n), is

Solution:

The angle of incidence for total polarization is given by tan θ = n ⇒ θ = tan -1 n
Where n is the refractive index of the glass.

QUESTION: 21

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is

Solution:

For constructive interference d sin θ = nλ

n = 0, 1,  – 1, 2,  – 2  hence five maxima are possible

QUESTION: 22

An electromagnetic wave of frequency n = 3.0 MHz passes from vacuum into a dielectric medium with permittivity ∈ = 4.0 . Then

Solution:

Frequency  remains constant during refraction

∴ wavelength is halved and frequency remains unchanged

QUESTION: 23

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

Solution:

QUESTION: 24

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of 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]

Solution:

QUESTION: 25

A thin glass (refractive index 1.5) lens has optical power of – 5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be

Solution:

.... (i)

... (ii)

QUESTION: 26

A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

Solution:

The shape of interference fringes formed on a screen in case of a monochromatic source is a straight line.
Remember for double hole experiment a hyperbola is generated.

QUESTION: 27

If I0 is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?

Solution:

When the slit width is doubled, the amplitude of the wave at the centre of the screen is doubled, so the intensity at the centre is increased by a factor 4.

QUESTION: 28

When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

Solution:

I = I0 cos2θ

Intensity of polarized light

⇒ Intensity of untransmitted ligh

QUESTION: 29

The refractive index of a glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviation for red and blue light respectively in a prism of this glass.
Then,

Solution:

For a thin prism, D = (μ – 1) A

Since λb < λr ⇒ μr < μb ⇒ D1 < D2

QUESTION: 30

In a Young’s double slit experiment the intensity at a point where the path difference is λ/6 (λ being the wavelength of light used) is I. If I0 denotes the maximum intensity,  equal to

Solution:

The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is φ is given by

is the maximum intensity..

NOTE : This formula is applicable when I1 = I2. Here

QUESTION: 31

Two lenses of power –15 D and +5 D are in contact with each other. The focal length of the combination is

Solution:

Power of combination is given by
P = P1 + P2 = (– 15 + 5) D = – 10 D.

QUESTION: 32

In an experiment, electrons are made to pass through a narrow slit of width ‘d’ comparable to their de Broglie wavelength.
They are detected on a screen at a distance ‘D’ from the slit (see figure).

Which of the following graphs can be expected to represent the number of electrons ‘N’ detected as a function of the detector position ‘y’(y = 0 corresponds to the middle of the slit)

Solution:

The electron beam will be diffracted and the maxima is obtained at y = 0. Also the distance between the first minima on both side will be greater than d.

QUESTION: 33

A student measures the focal length of a convex lens by putting an object pin at a distance ‘u’ from the lens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the student should look like

Solution:

This graph obeys the lens equation

where f is  a positive constant for a given convex lens.

QUESTION: 34

An experment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by

Solution:

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope

QUESTION: 35

A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on  the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light.
From this data, the wavelength of the unknown light is:

Solution:

Third bright fringe of known light coincides with the 4th bright fringe of the unknown light.

QUESTION: 36

A transparent solid cylindrical rod has a refractive index of  It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.

The incident angle θ for which the light ray grazes along the wall of the rod is :

Solution:

Applying Snell’s law at Q

Applying Snell’s Law at P

QUESTION: 37

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coordinates of P will be :

Solution:

Here u = -2 f , v = 2f

As |u| increases, v decreases for | u | >f. The graph between |v| and |u| is shown in the figure. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P (2f, 2f).

QUESTION: 38

An initially parallel cylindrical beam travels in a medium of refractive index μ (I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

Q. As the beam enters the medium , it will

Solution:

In the medium, the refractive index will decrease from the axis towards the periphery of the beam.
Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

QUESTION: 39

An initially parallel cylindrical beam travels in a medium of refractive index μ (I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

Q. The initial shape of the wavefront of the beam is

Solution:

Initially the parallel beam is cylindrical . Therefore, the wavefront will be planar.

QUESTION: 40

An initially parallel cylindrical beam travels in a medium of refractive index μ (I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

Q. The speed of light in the medium is

Solution:

The speed of light (c) in a medium of refractive index (μ) is given by

where c0 is the speed of light in vacuum

As I is decreasing with increasing radius, it is maximum on the axis of the beam. Therefore, c is minimum on the axis of the beam.

QUESTION: 41

Let the x-z plane be the boundary between two transparent media. Medium 1 in z ³ 0 has a refractive index of √2 and medium 2 with z < 0 has a refractive index of √3. A ray of light in medium 1 given by the vector  is incident on the plane of separation. The angle of refraction in medium 2 is:

Solution:

Angle of incidence is given by

∠ i = 60°

QUESTION: 42

This question has a paragraph followed by two statements, Statement – 1 and Statement – 2. Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement – 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π.
Statement – 2 : The centre of the interference pattern is dark.

Solution:

A phase change of π rad appears when the ray reflects at the glass-air interface. Also, the centre of the interference pattern is dark.

QUESTION: 43

A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is :

Solution:

From mirror formula

QUESTION: 44

An electromagnetic wave in vacuum has the electric and magnetic field  which are always perpendicular to each other. The direction of polarization is given by  that of wave propagation by

Solution:

∵ The E.M. wave are transverse in nature i.e.,

QUESTION: 45

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference φ is given by :

Solution:

Substituting in equation (1)

QUESTION: 46

An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

Solution:

The focal length of the lens

Now the object distance u.

QUESTION: 47

Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2×108 m/s, the focal length of the lens is

Solution:

∴ n = 3/2
32 + (R – 3mm)2 = R2

⇒  32 + R2 – 2R(3mm) + (3mm)2 = R2
⇒  R ≈ 15 cm

QUESTION: 48

Abeam of unpolarised light of intensity I0 is passed through a polaroidAand 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

Solution:

Relation between intensities

QUESTION: 49

Two coherent point sources S1 and S2 are separated by a small distance 'd' as shown. The fringes obtained on the screen will be

Solution:

It will be concentric circles.

QUESTION: 50

The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by

Solution:

For the prism as the angle of incidence (i) increases, the angle of deviation (δ) first decreases goes to minimum value and then increases.

QUESTION: 51

A thin convex lens made from crown glass  has focallength f. When it is measured in two different liquids having refractive indices  it has the focal lengths f1 and f2 respectively. The correct relation between the focal lengths is:

Solution:

By Lens maker's formula for convex lens

QUESTION: 52

A green light is incident from the water to the air - water interface at the critical angle (θ). Select the correct statement.

Solution:

For critical angle θc,

For greater wavelength or lesser frequency m is less.
So, critical angle would be more. So, they will not suffer reflection and come out at angles less then 90°.

QUESTION: 53

Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two  beams are IA and IB respectively, then

Solution:

According to malus law, intensity of emerging beam is given by,

I = I0cos2θ
Now, IA' = I A cos230º
IB ' = IB cos 60º
As IA ' = IB'

QUESTION: 54

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is :

Solution:

QUESTION: 55

On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground.
When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam:

Solution:

QUESTION: 56

Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is µ, a ray, incident at an angle θ, on the face AB would get transmitted through the face AC of the prism provided :

Solution:

When r2 = C, ∠N2Rc = 90°
Where C = critical angle

Applying snell's law at ‘R’
µ sin r2 = 1 sin90° ...(i)
Applying snell's law at ‘Q’
1 × sin θ = µ  sin r1 ...(ii)
But r1 = A – r2

So, sin θ =  µ  sin (A – r2)
sin θ = µ  sin A cos r2 – cos A         ...(iii)         [using (i)]
From (1)

...(iv)

By eq. (iii) and (iv)

on further solving we can show for ray not to transmitted through face AC

So, for transmission through face AC

QUESTION: 57

The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when :

Solution:

For b to be minimum

QUESTION: 58

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :

Solution:

A telescope magnifies by making the object appearing closer.

QUESTION: 59

In an experiment for determination of refractive index of glass of a prism by i – δ, plot it was found thata ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?

Solution:

We know that i + e -A = d

35° + 79° – A = 40°           ∴   A = 74°

μmax can be 5/3. That is μmax is less than 5/3 = 1.67

But δm will be less than 40° so