Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 72

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Question 1

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Gita arranges some mirrors in group- $$1$$ and group- $$2$$ according to the magnification, type of image and size of image from the following data.

Sr. No.	Group-$$1$$	Group-$$2$$
$$(1)$$	$$<1$$ and negative	Real, inverted and small
$$(2)$$	$$>1$$ and positive	Real, inverted and enlarged
$$(3)$$	$$>1$$ and negative	Virtual, erect and enlarged
$$(4)$$	$$<1$$ and positive	Virtual, erect and small

For which of the pairs from the group $$1$$ and $$2$$ you disagree.

$$1$$ and $$3$$

$$2$$ and $$3$$

$$3$$ and $$4$$

$$1$$ and $$2$$

Solution

If the magnification, $$m >1$$ and positive, then the image formed is virtual, erect and enlarged in nature. Also, when $$m>1$$ and negative, then the image formed is real, inverted and magnified in nature.

Question 2
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In a car a rear view mirror having a radius of curvature 1.50 m forms a virtual image of a bus located 10.0 m from the mirror. The factor by which the mirror magnifies the size of the bus is close to

A
0.06

B
0.07

C
0.08

D
0.09

Solution

The rear view mirror is a convex mirror.
Given:
Radius of curvature $$R=+1.50 \ m$$
Focal length $$f=\dfrac{R}{2}=\dfrac{+1.50}{2}=+0.75 \ m$$
Object distance $$u=-10 \ m$$

By mirror formula,
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\Rightarrow\dfrac{1}{0.75}=\dfrac{1}{v}+\dfrac{1}{-10}$$

$$\Rightarrow \dfrac{1}{v}=\dfrac{10+0.75}{7.5}=\dfrac{10.75}{7.5}$$
$$\Rightarrow v=\dfrac{750}{1075}=\dfrac{30}{43} \ m$$

Magnification $$m=-\dfrac{v}{u}=-\dfrac{30}{43} \times \dfrac{1}{(-10)}$$
$$\therefore m=+0.069 \approx +0.07$$

Option B is correct.
Question 3
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From the understanding of cartesian sign convention for reflection by spherical mirror, students took part in group discussion for $$FA-I$$ in classroom. Who is wrong in the group discussion?
Alpesh : All the distances are measured from the pole of a mirror parallel to the principal axis.
Beena: The distance measured in the direction of incident ray are taken positive.
Champak: The height measured upward and perpendicular to principal axis is taken negative.
Daksha : The height measured downward and perpendicular to principal axis is taken as positive.

A
Champak and Daksha

B
Only Champak

C
Alpesh and Beena

D
Only Daksha

Solution

The height measured upward and perpendicular to the principal axis is taken as positive. The height measured downward and perpendicular to the principal axis is taken as negative.
Question 4
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The magnification produced by a mirror is $$+\dfrac{1}{3}.$$ Then the mirror is a ____________.

A
Concave mirror

B
Convex mirror

C
Plane mirror

D
plano convex mirror

Solution

The value of $$m$$ is less than $$1$$. So it means the image is diminished.
$$m=+\dfrac{1}{3}$$
The positive sign indicates that the image is virtual and erect.

Hence, the mirror must be $$convex$$ mirror. Concave mirror also produces virtual image but it is enlarged. But convex mirror always produces diminished, virtual and erect image.

Answer-(B)
Question 5
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Choose the correct answer from the alternatives given.
There is certain material developed in laboratories which have a negative refractive index. A ray incident from the air (medium 1) into such a medium (medium 2) shall follow a path given by:

A

B

C

D

Solution

The negative refractive index materials are those in which incident rays from the air (medium 1) refracts or bends differently - opposite and symmetric to normal to that of positive refractive index medium.
Using Snell's law
$$\mu = \dfrac{sin \, i}{sin \, r} \Rightarrow sin \, r = \dfrac{sin \, i}{\mu}$$
According to the question, $$\mu$$ is negative
$$\therefore sin \, r$$ is negative.
Hence, r is negative
So light ray will bend as shown in figure
So the answer is (A)
Question 6
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The image of the needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. The displacement of the image, if the needle is moved by 5.0 cm away from the lens is

A
10 cm, toward the lens

B
15 cm, away from the lens

C
15 cm, towards the lens

D
10 cm, away from the lens

Solution

Here, u = -45 cm, v = 90 cm
$$\therefore$$ $$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{90} + \dfrac{1}{45} =\dfrac{1}{30} \Rightarrow f = 30 cm $$
when the needle is move 5 cm away from the lens,
u=-(45 + 5) = -50 cm
$$\therefore$$ $$\frac{1}{v'} = \dfrac{1}{f} +\dfrac {1}{u'} = \dfrac{1}{30} + \dfrac{1}{(-50)} = \dfrac{2}{150} = \Rightarrow v' = 75 cm $$
$$\therefore$$ Displacement of image = v-v' =90-75= 15 cm, towards the lens
Question 7
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A convex mirror forms an image one-fourth the size of the object. If object is at a distance of $$0.5\ m$$ from mirror the focal length of the mirror is

A
$$0.16\ m$$

B
$$-1.5\ m$$

C
$$0.4\ m$$

D
$$-0.4\ m$$

Solution

Given :    $$u = -0.5 \ m$$
Convex mirror forms erect image of real object.
Using,   $$\dfrac{h_i}{h_o} = -\dfrac{v}{u}$$
Or   $$\dfrac{1}{4} = -\dfrac{v}{(-0.5)}$$
$$\implies \ v = 0.125 \ m$$
Using mirror formula,   $$\dfrac{1}{v}+\dfrac{1}{u} = \dfrac{1}{f}$$
$$\therefore$$   $$\dfrac{1}{0.125}+\dfrac{1}{-0.5} = \dfrac{1}{f}$$
$$\implies \ f = 0.16 \ m$$
Question 8
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A convex mirror of focal length $$f$$ forms an image which is $$\cfrac { 1 }{ n } $$ times the object. The distance of the object from the mirror is :

A
$$\left( \cfrac { n-1 }{ n } \right) f$$

B
$$\left( \cfrac { n+1 }{ n } \right) f$$

C
$$(n+1)f$$

D
$$(n-1)f$$

Solution

Given,

Magnification, $$m=\cfrac { 1 }{ n } $$

So, $$\cfrac { 1 }{ n } =-\cfrac { v }{ u } $$

or $$v=-\cfrac { u }{ n } $$

By using mirror formula, $$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } +\cfrac { 1 }{ u } $$

$$=\cfrac { 1 }{ -u/n } +\cfrac { 1 }{ u } $$

$$\Rightarrow u=-(n-1)f$$
Question 9
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Magnification produced is +$$\dfrac { 1 }{ 3 }$$, then what kind of mirror it is?

A
concave mirror

B
convex mirror

C
opaque mirror

D
plane mirror

Solution

Since the magnification produced is positive and less than 1, it means the image is erect and diminished. This kind of image is possible using a convex mirror.
Question 10
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A ray $$R_{1}$$ is incident on the plane surface of the glass slab (kept in air) of refractive index $$\sqrt {2}$$ at an angle of incidence equal to the critical angle for this air glass system. The refracted ray $$R_{2}$$ undergoes partial reflection and refraction at the other surface. The angle between reflected ray $$R_{3}$$ and the refracted ray $$R_{4}$$ at that surface is :

A
$$45^{\circ}$$

B
$$135^{\circ}$$

C
$$105^{\circ}$$

D
$$75^{\circ}$$

Solution

Since, it is given that angle of incidence is equal to the critical angle

Therefore,

$$ \sin \,i\,=\,\dfrac{1}{n}\,\,=\,\dfrac{1}{\sqrt{2}} $$

$$ i\,=\,\,45{}^\circ $$

$$ $$According to laws of refraction

$$ \dfrac{\sin i}{\sin r}\,=\,n\,\,(refractive\,index) $$

$$ sin\,r\,=\,\dfrac{\sin \,i}{n}\,=\,\,\frac{1}{2} $$

$$ r\,=\,30{}^\circ $$

Let $$\theta $$ is the angle between R₃and R₄

So, $$ \theta \,=\,180{}^\circ \,-\,(i\,+\,r)\, $$

$$ \,\,\,\,\,=\,180\,-\,(\,45\,+\,30\,) $$

$$ \,\,\,\,\,\,=\,\,105{}^\circ $$