Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 40

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Question 1
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An object of length $$4 \ cm$$ is placed in front of a concave mirror at a distance of $$30 \ cm$$. The focal length of mirror is $$15\ cm$$. What will be the length of image?

A
$$1 \ cm$$

B
$$2 \ cm$$

C
$$4 \ cm$$

D
$$6 \ cm$$

Solution

The relationship between the radius of curvature and focal length is that the radius of curvature is twice the focal length.

In this case, focal length of the concave mirror is given as $$15\ cm$$. Hence, the radius of curvature is $$15\times 2=30\ cm$$.

When the object is placed on the centre of curvature, on the principal axis, in front of a concave lens, the image is formed at the same point. The image formed is real and inverted.

So, in this case, the object distance is equal to the image distance.
Therefore, the object size is equal to the image size.

As the object is of $$4\ cm$$ in length, the image will also be of $$4\ cm$$ in length.
Question 2
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A ray of light travelling inside a rectangular glass block of refractive index $$\sqrt2$$ is incident on the glass-air surface at an angle o incidence of $$45^o$$.The refractive index of air is one. Under these conditions the ray will

A
emerge into the air without any deviation

B
be reflectived back into the glass

C
be absorbed

D
emerge into the air with an angle of refraction equal to $$90_o$$

Solution

Using Snell's law:

$${ \mu }_{ i }\sin { \left( i \right) } ={ \mu }_{ r }\sin { \left( r \right) } $$

$$\Rightarrow \sqrt { 2 } \sin { \left( 45 \right) } =\sin { \left( r \right) } \\ \therefore \quad r=90$$

therefore, the ray will emerge into air at $$90^0$$ , this angle of incidence is called as critical angle.
Question 3
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An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20 cm. The image produced is:

A
real, inverted and smaller in size

B
real, inverted, and of same size

C
real and erect

D
virtual and inverted

Solution

Object distance, $$u = -40 cm$$

Focal length, $$f =-20 cm$$

According to mirror formula,

$$\displaystyle \frac {1}{u} + \frac {1}{v} = \frac {1}{f}\ \ \ or\ \ \ \frac {1}{v} = \frac {1}{f} - \frac {1}{u}$$

or $$\displaystyle \frac {1}{v} + \frac {1}{-20} - \frac {1}{(-40)} = \frac {1}{-20} + \frac {1}{40}$$

$$\displaystyle \frac {1}{v} =\frac {-2+1}{40} = -\frac{1}{40}\ \ or\ \ \ v = -40 cm$$

Negative sign shows that image is infront of concave mirror.
The image is real.

Magnification, $$\displaystyle m = \frac {-v}{u} = -\frac {(-40)}{(-40)}= -1$$

The image is of the same size and inverted.
Question 4
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The path of a ray of light coming from air passing through a rectangular slab traced by four students are shown by figures A, B, C and D. Which one of them is correct?

A

B

C

D

Solution

Answer is B.

When light travels from one medium to another, it generally bends, or refracts. The law of refraction gives us a way of predicting the amount of bend. The law of refraction is also known as Snell's law, named for Willobrord Snell, who discovered the law in 1621.
Snell's law gives the relationship between angles of incidence and refraction for a wave impinging on an interface between two media with different index of refraction.
Thus, when a ray of light passes through the rectangular slab, the angle of incidence is equal to angle of emergence and this implies that, the incident ray and the emergent ray are parallel to each other.
Hence, option B is correct.
Question 5
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After tracing the path of a ray of light passing through a rectangular glass slab for four different values of the angle of incidence, a student reported his observations in tabular form as given below :

S.No <i <r < e

I $${30^0}$$ $${19^0}$$ $${29^0}$$
II $${40^0}$$ $${28^0}$$ $${40^0}$$
III $${50^0}$$ $${36^0}$$ $${50^0}$$
IV $${60^0}$$ $${40^0}$$ $${59^0}$$
The best observation is :

A
I

B
II

C
III

D
IV

Solution

Observation IV will give the best result because it has the largest angle of incidence, due to which the lateral displacement between the incident ray and the emergent ray will be maximum for light rays passing through a glass slab. Hence, the correct option is D.
Question 6
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In the glass slab experiment,the correct placement of the protractor (or 'dec'),for measuring the angles of incidence and emergence,is shown in which figure?

A

B

C

D

Solution

The protractor should always be placed on such a way that the base is always along the normal of the incident ray or the emergent ray.
Hence, option D is the correct way.
Question 7
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A point source of light is placed in front of a plane mirror, then :

A
all the reflected rays will meet at a point when produced backward

B
only the reflected rays close to the normal will meet at a point when produced backward

C
only the reflected rays making a small angle with the mirror will meet at a point when produced backward

D
light of different colours will make different images

Solution

The image of a point source in plane mirror will be a virtual point image, behind the mirror. So the reflected rays should meet at this point when produced backwards.
Question 8
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A printed page is seen through a glass slab placed on it. The printed words appear raised. This is due to :

A
refraction at the upper surface of the slab

B
refraction at the lower surface of the slab

C
partial refraction at the upper surface of the slab

D
partial refraction at the lower surface of the slab

Solution

This is due to Refraction at the upper surface of the slab. In this case, the object is in the denser medium and the observer is in the rarer medium. So, due to refraction, the apparent depth of the printed page will be a little less than the real depth or actual depth.
Question 9
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Which of the following (referred to as spherical mirror) depends on whether the rays are paraxial or not?

A
Pole

B
Focus

C
Radius of curvature

D
Principal axis

Solution

The pole, the radius of curvature, and the principal axis are the geometric nomenclatures of points and axis respectively.
However focus is the only point where a parallel beam of light concentrates after reflection, which depends upon the nature of rays.
Question 10
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An incident ray passing through the focus will be reflected :

A
straight away

B
through the pole of the mirror

C
back and passes through the centre of curvature

D
parallel to the principal axis

Solution

An incident ray passing through the focus will move parallel to the principal axis after reflection from spherical mirror.