Ray Optics and Optical Test

Result

Ray Optics and Optical Test - 55

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A point object is placed on principal axis of concave mirror of radius of curvature 10 cm at a distance 21 cm from pole of the mirror. A glass slab of thickness 3 cm and refractive index 1.5 is placed between object and mirror $$.$$ Find the imaged position of the image formed in cm is

A
4

B
3

C
7.67

D
5

Solution
Question 2
1 / -0

Which colour shows maximum deviation when light is dispersed through prism?

A
Violet

B
red

C
yellow

D
blue
Question 3
1 / -0

An object is placed $$12 \ cm$$ to the left of a converging lens of focal length $$8 \ cm $$. Another converging lens of $$6 \ cm$$ focal length is placed at a distance of $$ 30 \ cm$$ to the right of the first lens. The second lens will produce

A
No image

B
A virtual enlarged image

C
A real enlarged imnage

D
a raeal inverted inage.
Question 4
1 / -0

Three right angied prisms of refractive indices $$\mu_1$$,$$\mu_2$$ and $$\mu_3$$ are joined together so that the faces of the middle prism are each in contact with one the outside prism.If the ray passes through the composite block undeviated,then

A
$${\mu_1}^2+{\mu_3}^2-{\mu_2}^2 = 1$$

B
$${\mu_1}^2-{\mu_3}^2+{\mu_2}^2 = 1$$

C
$${\mu_1}^2-{\mu_3}^2-{\mu_2}^2 = 1$$

D
$${\mu_2}^2+{\mu_3}^2-{\mu_1}^2 = 1$$

Solution
Question 5
1 / -0

The angle of minimum deviation measured with a prism is $$30^o$$ and the angle of prism $$60^o$$. The refractive index of prism material is -

A
$$\sqrt{2}$$

B
$$2$$

C
$$\dfrac{3}{2}$$

D
$$\dfrac{4}{3}$$

Solution

As we know,
$$\mu=\dfrac{\sin \left(\dfrac{A+\delta_{m}}{2}\right)}{\sin (A/2)}=\dfrac{\sin 45^{o}}{\sin 30^{o}}=\sqrt{2}$$
Question 6
1 / -0

When light rays are incident on a prism at an angle of $$45^o$$, the minimum deviation is obtained. If the refractive index of the material of the prism is $$\sqrt 3$$, then the angle of prism will be

A
$$2 \times {\sin ^{ - 1}}\frac{1}{{\sqrt 6 }}$$

B
$$45^o$$

C
$$60^o$$

D
$$90^o$$

Solution

Given
$$\theta=45^\circ$$
$$\mu=\sqrt3$$

$$\because A=r_1+r_2$$

For minimum deviation, $$r_1=r_2$$

$$r_1=r_2=\dfrac A2$$

From Snell's law
$$\mu_1\sin\theta=\mu_2\sin r_1$$

$$1\cdot\sin45=\sqrt3\ \sin r_1$$

$$\sin r_1=\dfrac1{\sqrt6}$$

$$r_1=\sin^{-1}\dfrac1{\sqrt6}$$

$$A=2r_1=2\sin^{-1}\dfrac1{\sqrt6}$$
Option A
Question 7
1 / -0

A thin lens produce an image of the same size as the object.Then first the optical center of the lens, distance of the object is

A
Zero

B
4f

C
2f

D
f/2

Solution

A thin lens produce an image of the same size as the object.Then first the optical center of the lens, distance of the object is 2f.
so that the correct option is C.
Question 8
1 / -0

Three right angled prisms of refractive indices $$\mu_1$$, $$\mu_2$$ and $$\mu_3$$ are joined together ao that the faces of the middle prism are each in contact with one of the outside prisms. If the ray passes through the composite block undeviated, then

A
$$1 + \mu _2^2 = \mu _1^2 + \mu _3^2$$

B
$${\mu_1}^2-{\mu_3}^2+{\mu_2}^3 = 1$$

C
$${\mu_1}^2-{\mu_3}^2-{\mu_2}^3 = 1$$

D
$${\mu_2}^2+{\mu_3}^3-{\mu_2}^1 = 1$$

Solution

$$\alpha = 90-r_i$$
$$\beta = 90-r_2$$
$$\gamma = 90-r_3$$
$$\begin{array}{l} Applying\, snell's\, Law\, at\, each\, surface \\ AtB, \\ \sin i={ \mu _{ 1 } }\sin { r_{ 1 } } \Rightarrow { \sin ^{ 2 } }i=\mu _{ 1 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 1 } } \\ At\, C \\ { \mu _{ 1 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 2 } }\sin { r_{ 2 } } \Rightarrow \mu _{ 1 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 2 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 2 } } \\ At\, D \\ { \mu _{ 2 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 3 } }\sin { r_{ 3 } } \Rightarrow \mu _{ 2 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 3 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 3 } } \\ At\, E, \\ { n_{ 3 } }\sin \left( { 90-{ r_{ 3 } } } \right) =\sin \left( { 90-i } \right) \Rightarrow \mu _{ 3 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 3 } }={ \cos ^{ 2 } }i \\ Adding\, all\, above\, equations\, \, gives, \\ 1+\mu _{ 2 }^{ 2 }=\mu _{ 1 }^{ 2 }+\mu _{ 3 }^{ 2 } \\ Hence,\, the\, option\, \, A\, is\, the\, correct\, answer. \end{array}$$
Question 9
1 / -0

A thick plane mirror shows a number of images of the filament of an electric bulb. Of these, the brightest image is the:

A
First

B
Second

C
Last

D
Fourth

Solution

A thick plane mirror forms a number of images of the filament of an electric bulb. Images formed by the front surface which is unpolished reflect only a small part of the light (say $$10\% $$), while the second image is formed by the polished surface which reflects most of the intensity ($$90\%). Hence the second image is the brightest. The intensity of the next reflections will keep on decreasing.
Question 10
1 / -0

A coordinate axis as shown in the the figure is kept in front of a converging lens at a distance 2f from it, where f is the focal length of the lens. Which of the following shows the approximate shape of the image? Assume that X-axis is the principal axis of the lens.

A

B

C

D

Solution