Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 39

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Question 1
1 / -0

During refraction through a glass block, the angle of refraction _______.

A
increases

B
decreases

C
remains same

D
cant say

Solution

Can't say as angle of refraction depends upon angle of incidence.
Question 2
1 / -0

The refractive index of glass is $$1.5$$. From a point $$P$$ inside a glass block, draw rays $$PA, PB$$ and $$PC$$ incident on the glass-air surface at an angle of incidence $$30^o, 42^o$$ and $$60^o$$ respectively. What is the angle of refraction for the ray $$PB$$ ? (Take $$\sin 42^o = \cfrac{2}{3}$$ )

A
$$0^o$$

B
$$30^o$$

C
$$60^o$$

D
$$90^o$$

Solution

According to Snell's Law: $$\cfrac{sin \ i}{sin \ r}=\cfrac{n_2}{n_1}$$
Substituting, $$\cfrac{sin \ 42^o}{sin \ r}=\cfrac{1}{1.5}$$
Thus, $$sin \ r=\cfrac{1.5}{1.5}=1$$
Thus, $$r=90^o$$
Question 3
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The centre of curvature of a concave mirror :

A
lies in front of it

B
lies behind it

C
lies on the surface of mirror

D
none of these

Solution

From the attached figure, it can be observed that the centre of curvature C lies in front of the concave mirror.
Question 4
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Directions For Questions

In the diagram, $$M$$ is a concave mirror, and $$A$$ is a point on its principal axis. If an object $$O$$ is kept at $$A$$, the image is formed at $$A$$ itself. B is the center of AP.

...view full instructions

What is the distance $$PB$$ called?

A
Efficient length

B
Radius of curvature

C
Equivalent length

D
Focal length.

Solution

In case of concave mirror, the position of image and object will coincide if and only if the object is placed at the center of curvature of mirror.

Hence, PA is equal to the distance of centre of curvature from the pole = $$R$$
Now, If B is the center of PA, then PB will be equal to focal length.
Since, focal length $$f=\dfrac{R}{2}$$

Hence, distance PB is the focal length of the mirror.
Question 5
1 / -0

According to the new Cartesian sign convention, the _______ of a curved mirror is taken at origin.

A
Centre of curvature

B
Pole of the mirror

C
Focus

D
Midpoint of focus and pole

Solution

In the case of mirrors, the pole(optic center) is the origin to measure object and image distances.
Distances that are measured in the direction of the incident rays are positive while distances measured in the direction opposite to the incident rays are taken to be negative.
Question 6
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Which of the following ray diagram is correct ?

A

B

C

D

Solution

A ray parallel to principal axis must pass through focus and a ray passing through focus must go parallel to principal axis. Hence, the option A is correct.
Question 7
1 / -0

A concave lens has focal length of $$15 \ cm$$. At what distance should the object from the lens be placed so that it forms an erect and virtual image at $$10 \ cm$$ from the lens?

A
$$\displaystyle 30$$ $$cm$$

B
$$\displaystyle 15$$ $$cm$$

C
$$\displaystyle 60$$ $$cm$$

D
$$\displaystyle 10$$ $$cm$$

Solution

Given:
Focal length $$ f = -15 cm$$
Image distance $$v = -10 cm$$
Using lens formula,

$$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \Rightarrow \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}$$

$$\Rightarrow \dfrac{1}{u} = -\dfrac{1}{10} - \left ( \dfrac{1}{-15} \right )$$

$$ = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-3 + 2}{30}$$

$$\Rightarrow u = -30 cm$$

Answer is option A.
Question 8
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For an object placed at a distance $$20 \ cm$$ in front of a convex lens, the image is at a distance $$20 \ cm$$ behind the lens. The focal length of convex lens is:

A
$$20 \ cm$$

B
$$10 \ cm$$

C
$$15 \ cm$$

D
$$40 \ cm$$

Solution

Answer is B.

There is a geometrical relationship between the focal length of a lens $$f$$, the distance from the lens to the bright object $$o$$ and the distance from the lens to the projected image $$i$$. The relationship between the distances is given as $$\dfrac { 1 }{ i } -\dfrac { 1 }{ o } =\dfrac { 1 }{ f } $$.
It is given that object distance is -$$20$$ cm and the image distance is $$20 cm$$. So, the focal length is given as $$\dfrac { 1 }{ 20 } +\dfrac { 1 }{ 20 } =\dfrac { 1 }{ 0.1 } $$. The focal length $$f=\dfrac { 1 }{ 0.1 } = 10 cm.$$

The focal length of the lens is 10 cm.
Question 9
1 / -0

Directions For Questions

In the diagram, $$M$$ is a concave mirror, and $$A$$ is a point on its principal axis. If an object $$O$$ is kept at $$A$$, the image is formed at $$A$$ itself. B is the center of AP.

...view full instructions

What is the distance $$PA$$ called?

A
Equivalent distance

B
Equivalent radius

C
Radius of curvature

D
Center of curvature

Solution

When the image forms on the object in a concave mirror, then the object as well as the image is situated at the center of the curvature of the concave mirror. Hence, the distance $$PA$$ is called as Radius of Curvature of the concave mirror.
Relation between radius of curveture(R) and focal length(f)-
$$f = \dfrac{ R }{ 2 }$$
$$PB = \dfrac{PA}{2}$$
Question 10
1 / -0

A ray of light is incident on a lens parallel to its principal axis. After refraction, the ray passes through or appear to come from :

A
its first focus

B
its optical centre

C
its second focus

D
the centre of curvature of its second surface

Solution

The parallel rays of light meet at a point on the principal axis after refraction by a convex lens or appear to come from a point on the principal axis after refraction by a concave lens. This point is known as the second focus of the lens.