Question Set B

cbse 12th Standard Physics Wave Optics Model Questions Papers Study Material - QB365

Download CBSE Class 12th Standard CBSE Physics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Physics

Questions:

Generic 2

18 x 2 = 36

A convex lens of local length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature ?

The radii of curvature of the face of a double convex lens are 10 cm and 15 cm. If focal length of the length is 12 cm, find the refractive index of the material of the lens.

What is the value of refractive index of a medium of polarising angle \({ 60 }^{ ° }\)?

How does the resolving power of a microscope change on
(i) decreasing wavelength of light
(ii) decreasing diameter of objective lens?

What happens to light energy when light waves interfere destructively at a point?

A beam of light is converging towards a certain point. A parallel sided glass plate is introduced in the path of the converging beam.How will the point of convergence be shifted?

Define Snell's law of refraction.

State the factors on which refractive index of a medium depend?

How can we see a virtual image when it cannot be obtained on a screen?

10.

If the wavelength of incident light on a concave mirror is increased, how will the focal length of the mirror change?

11.

why does Galilean telescope have a smaller field of view?

12.

Can a virtual image be photographed?

13.

Why is the aperture of objective lens of a telescope taken large?

14.

An astronomical telescope uses lenses of power 10D and 1D.What is it its magnifying power in mnormal adjustment?

15.

A mirror is turned through\(15°\). Through what angle will the reflected ray turn?

16.

Define principle axis of a spherical mirror.

17.

(a) giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eye-piece of focal length 15 m. If an eye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
(b) If this telescope is used to view the moon, what is the diameter of the image od the moon formed by the objective lens? The diameter of the moon id 3.48 \(\times\) 10⁶ m, and the radius of lunar is 3.8 \(\times\)10⁸m.

18.

A compound microscope consists of an objective lens of focal length 2.0 cm and an eye -piece of focal length 6.25 cm separated by a distance should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), (b) infinity ? What is the magnifying power of the microscope in each case ?

Generic 3

3 x 3 = 9

19.

Light of wavelength \(6000\overset { \circ }{ A } \) falls on a plane reflecting surface.What are the wavelength and frequency of reflected light. If angle between incident ray and reflected ray is \(60° \), what is the angle of incidence?

20.

If two slits in Young's double slit experiment are having width ratio 4:9, then find the ratio of intensity at maxima to the intensity at minima.

21.

(a) For the telescope described in - What is the separation between the objective lens and eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens
(c) What is the height of the final image on the tower if it is formed at 25 cm?

Generic 5

3 x 5 = 15

22.

Define magnifying power of a telescope. Write its expression. A small telscope has an objective lens of focal length 150cm and an eyepiece of focal length 5cm. If this telescope is used to view a 100m high tower 3Km away, find the height of the final image, when it is formed 25cm away from the eyepiece.

23.

The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and eye piece is 101 cm. Calculate the focal lengths of objective and eye piece.

24.

Distinguish between interference and diffraction.

Multiple Choice Question

10 x 1 = 10

25.

The angle between pass axis of polarizer and analyzer is \(45°\)The percentage of polarized light passing through analyzer is

(a)

100%

(b)

50%

(c)

25%

(d)

75%

26.

Which of the following cannot be polarized?

(a)

X-rays

(b)

radio waves

(c)

sound waves

(d)

light waves

27.

Image of an object in a concave mirror is

(a)

always real

(b)

always virtual

(c)

always erect

(d)

real or virtual depending on position of object

28.

For any position of an object, image formed in a convex mirror is

(a)

virtual

(b)

erect

(c)

smaller in size

(d)

as far behind the mirror as the object is in front

29.

The correct mirror equation is

(a)

\(\frac { 1 }{ f } =\frac { 1 }{ \upsilon } +\frac { 1 }{ u } \)

(b)

\(\frac { 1 }{ f } =\frac { 1 }{ \upsilon } -\frac { 1 }{ u } \)

(c)

\(\frac { 1 }{ f } =\frac { 1 }{ u } -\frac { 1 }{ \upsilon } \)

(d)

none of these

30.

The relation between focal length \(f\) and radius of curvature \(R\) of a spherical mirror is

(a)

\(f=R\)

(b)

\(f=R/2\)

(c)

\(f=2 R\)

(d)

none of these

31.

Which is not true for the image formed in a plane mirror? The image is

(a)

virtual

(b)

erect

(c)

laterally inverted

(d)

closer to the mirror than the object

32.

In a concave mirror, an object is placed at a distance \({ d }_{ 1 }\) from the focus and the real image is formed at a distance \({ d }_{ 2 }\) from the focus. then the focal length of the mirror is:

(a)

\(\sqrt { { d }_{ 1 }{ d }_{ 2 } } \)

(b)

\({ d }_{ 1 }{ d }_{ 2 }\)

(c)

\({ { (d }_{ 1 }{ /d }_{ 2 }) }^{ /2 }\)

(d)

\(\sqrt { { d }_{ 1 }{ /d }_{ 2 } } \)

33.

The ratio of the speed of an object to the speed of its real image of magnification m in the case of a convex mirror is

(a)

\(-\frac { 1 }{ { m }^{ 2 } } \)

(b)

\({ m }^{ 2 }\)

(c)

-xm

(d)

\(\frac { 1 }{ { m } } \)

34.

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

(a)

The size decrease

(b)

The intensity increase

(c)

The size increase

(d)

The intensity decrease

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

18 x 2 = 36

Power of convex lens = 1/0.25 = 4D
Power of concave lens = 1/0.20 = -5D
Power of the combination, P = P₁+P₂ = -1D
Nature : Diverging

R₁= 10 cm, R₂ = -15 cm, f = 12 cm
\(\frac { 1 }{ f } =(\mu -1)(\frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } } )\)
\(\frac { 1 }{ 12 } =(\mu -1)(\frac { 1 }{ 10 } -\frac { 1 }{ 15 } )\)
\( \mu = \frac { 3 }{ 2 } \)

According to Brewster's law,
\(\mu =\tan { { i }_{ p } } =\tan { { 60 }^{ ° } } =\sqrt { 3 } =1.732\)

On decreasing \(\lambda \), resolving power of microscope increases and on decreasing diameter of objective lens, resolving power of microscope decreases.

Energy is not lost. It gets transferred from regions of destructive interference to the regions of constructive interference.

The point of convergence will be shifted away from the glass plate.

\(\mu =\frac { \sin { i } }{ \sin { r } } \), where symbols have their usual meaning.

Refractive index of a medium depends on
(i) nature of medium
(ii) wavelength of light used.
(iii) temperature
(iv) nature of surrounding medium

True, a virtual image cannot be taken on screen.But our eye lens forms a real image of(virtual image acting as virtual object) on the retina of our eyes.

10.

The focal length of mirror does not change on changing the wavelength of light incident on it. This is because f = R/2, i.e. f depends only on radius of curvature of the mirror.

11.

This is because eye lens in such a telescope is concave

12.

Yes, it can be photographed.

13.

This is done to increase the light gathering capacity and hence brightness of image.

14.

\(m={f_0\over f_e}={P_e\over P_o}={10\over 1}\)

15.

30^o as the reflected ray turns through twice the angle through which mirror is turned.

16.

A line joining the pole P to centre of curvature of the mirror, product on both sides is called principle axis of mirror.

17.

1500
13.7 cm

18.

20; 13.5 cm

Generic 3

3 x 3 = 9

19.

As wavelength and frequency remain unchanged on reflection, therefore,
\({ \lambda }^{ ' }=\lambda =6000\overset { \circ }{ A } \)
\({ v }^{ ' }=v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 6000\times { 10 }^{ -10 } } =5\times { 10 }^{ 14 }Hz\)
now,
\({ v }^{ ' }=v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 6000\times { 10 }^{ -10 } } =5\times { 10 }^{ 14 }Hz\)
\( \angle i+\angle r=60°\)
\( \angle r=\angle i \ \therefore 2\angle i=60°\)
\( i=30°\)

20.

\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } =\frac { 4 }{ 9 } or\frac { { a }^{ 2 } }{ { b }^{ 2 } } =\frac { 4 }{ 9 } \)
\( \ \frac { a }{ b } =\frac { 2 }{ 3 } or \ a=\frac { 2 }{ 3 } b\)
\( \frac { { I }_{ max } }{ { I }_{ min } } =\frac { { (a+b) }^{ 2 } }{ { (a-b) }^{ 2 } } =\frac { { (\frac { 2 }{ 3 } b+b) }^{ 2 } }{ { (\frac { 2 }{ 3 } b-b) }^{ 2 } } \)
\( \frac { { I }_{ max } }{ { I }_{ min } } =\frac { { (\frac { 5 }{ 3 } b) }^{ 2 } }{ { (-\frac { 1 }{ 3 } b) }^{ 2 } } =\frac { 25 }{ 1 } \)

21.

\((a)\quad { f }_{ 0 }+{ f }_{ e }=140+5=145 \ cm\)
\( (b)\ Angle \ subtended \ by \ the \ tower\ \)
\(=\frac { 100 }{ 3000 } =\frac { 1 }{ 30 } rad\quad \quad \quad \quad ....(i)\)
\(\\ =\frac { h }{ { f }_{ 0 } } =\frac { h }{ 140 } ....(ii)\)
\(\\ Equating(i)and(ii),weget\quad \)
\(\\ \frac { h }{ 140 } =\frac { 1 }{ 30 } \quad \quad \quad \quad\)
\( \\ h=\frac { 14 }{ 3 } \quad cm=4.7\quad cm\)
\(\\ (c)\quad Magnification(magnitude)oftheeyepiece\)
\( =1+\frac { D }{ f } =1+\frac { 25 }{ 5 } =6\)
\(\\ Heightofthefinalimage(magnitude)\)
\( =\frac { 14 }{ 3 } \times 6=28 cm\)

Generic 5

3 x 5 = 15

22.

The magnifying power of a telescope is equal to the ratio of the visual angle substended at the eye by final image formed at least distance of distinct vision to the visual angle subtended at naked eye by the object at infinity.

23.

m = -100, f₀+ f_e= 101 cm, f₀ = ?, f_e = ?
\(m=-{f_0\over f_e}=-100\ \ \therefore f_0=100 f_3\)
Now f₀ + f_e = 101
100 f_e + f_e= 101,
f_e = 1 cm,
f₀ = 100 f_e= 100 cm

24.

Differences between interference and diffraction

Interference	Diffraction
1. Interference takes place when light from two different wavefronts coming from two coherent sources superimpose on each other.	1. Diffraction is due to superposition of secondary wavelets from various points on the same wave-front.
2. Bright fringes are of the same intensity.	2. Intensity of secondary maximas goes on decreasing.
3. Fringes are equispaced	3. Fringes are not equispaced.
4. Intensity of light is zero at minima.	4. Intensity of light at minima is not zero.