Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 58

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Question 1
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Refraction of light occurs when:

A
light passes through a narrow opening.

B
light passes into a new transparent material (not head on) in which it travels at a new speed.

C
light strikes the edge of an object.

D
light becomes polarized.

E
light passes through a diffraction grating.

Solution

When light ray goes from one transparent medium to another transparent medium it changes its path and travels with a new speed , this phenomenon is called refraction .
Question 2
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A light ray travels at some angle (not head-on) from water in a rectangular aquarium, through the glass side, and out into the air. The indices of refraction for water, glass, and air are 1.33, about 1.6, and 1.0, respectively.
Rank the materials in descending order according to the angles of refraction made by the light ray with the normal in each.

A
air, water, glass.

B
air, glass, water.

C
water, air, glass.

D
water, glass, air

E
glass, water, air

Solution

From snell's law, for light travelling from one medium to another, $$\mu sinr=constant$$
where $$r$$ is the angle of refraction that the light makes with the normal in the medium,
$$\implies sin r\propto \dfrac{1}{\mu}$$
Since $$\mu_{glass}>\mu_{water}>\mu_{air}$$,
$$r_{air}>r_{water}>r_{glass}$$
Question 3
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Light moves from air into water. The light makes an angle of $$47^0$$ with the normal in the air. What angle will it make with the normal in the water?
The index of refraction for air is $$1.00$$.
The index of refraction for water is $$1.33$$.

A
$$47^o$$

B
$$77^o$$

C
$$43^o$$

D
$$25^o$$

E
$$33^o$$

Solution

Given : $$n_{air} = 1.00$$ $$n_{water} = 1.33$$
Angle of incidence $$\theta_i = 47^o$$ $$(sin47^o = 0.73)$$
Using Snell's law of refraction : $$n_{air} \times sin\theta_i = n_{water} \times sin\theta_r$$
$$\therefore$$ $$1.00 \times sin47^o =1.33 \times sin\theta_r$$ $$\implies sin\theta_r = 0.55$$ OR $$\theta_r \approx 33^o$$
Question 4
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A student traces the path of a ray of light through a rectangular glass slab for the different values of angle of incidence. He observes all possible precautions at each step of the experiment. At the end of the experiment, on analyzing the measurements, which of the following conclusions is he likely to draw?

A
$$\angle i = \angle e < \angle r$$

B
$$\angle i < \angle e < \angle r$$

C
$$\angle i > \angle e > \angle r$$

D
$$\angle i = \angle e > \angle r$$

Solution

As the incident ray is parallel to the emergent ray, thus incidence angle is equal to angle of emergence i.e $$\angle i = \angle e$$
Also the ray travelling from rarer medium (air) to the denser medium (glass slab) slightly bends towards the normal after refraction, thus angle of refraction is smaller than angle of incidence i.e $$\angle r < \angle i$$
$$\implies$$ $$\angle i = \angle e > \angle r$$
Question 5
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An object is placed somewhere in front of concave mirror. The focal length of the mirror is $$10.0 cm$$. The image for this object CANNOT be formed in which of the following locations?

A
$$25.0 cm$$ in front of the mirror

B
$$20.0 cm$$ in front of the mirror

C
$$15.0 cm$$ in front of the mirror

D
$$5.0 cm$$ in front of the mirror

E
$$10.0 cm$$ behind the mirror

Solution

Given : $$f = -10$$ cm
Let the object is placed at a distance $$x$$ from the pole of mirror i.e $$u = -x$$.
Using mirror formula : $$\dfrac{1}{v}+ \dfrac{1}{u} = \dfrac{1}{f}$$
$$\therefore$$ $$\dfrac{1}{v}+ \dfrac{1}{-x} = \dfrac{1}{-10}$$ $$\implies v = \dfrac{10x}{10-x}$$
For option D : $$v = -5.0$$ cm
$$\therefore$$ $$-5.0 = \dfrac{10 x}{10-x}$$ $$\implies x = -10$$ cm (not possible)
Hence option D is correct.
Question 6
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Material $$1$$ has an index of refraction of $$1.15$$.
Material $$2$$ has an index of refraction of $$2.30$$.
If light passes from air into each of these materials at the same angle of incidence, how will the angle of refraction in material $$1$$ compare to the angle of refraction in material $$2$$?

A
The angle of refraction in material $$1$$ will be half as much as the angle of refraction in material $$2$$

B
The angle of refraction in material $$1$$ will be greater than the angle of refraction in material $$2$$

C
The angle of refraction in material $$1$$ will be less than the angle of refraction in material $$2$$ but not necessarily half as much as the angle of refraction in material $$2$$

D
The angle of refraction in material $$1$$ will be less than the angle of refraction in material $$2$$ but not necessarily twice as much as the angle of refraction in material $$2$$

E
The angle of refraction in material $$1$$ will be one-fourth as much as the angle of refraction in material $$2$$

Solution

$$n_1$$=$$1.15$$ $$n_2$$ =$$2.30$$
Given : $$n_1< n_2$$
Refractive index of air $$n_a = 1.00$$
Using Snell's law of refraction, $$n_{air} \times sin i = n_{medium} \times sin r$$
$$\therefore$$ $$1.00 \times sin i = n_{medium} \times sin r$$ $$\implies $$ $$sin r = \dfrac{sin i}{n_{medium}}$$

Hence $$r$$ is inversely proportional to refractive index $$n$$
Thus $$n_1< n_2$$ $$\implies$$ $$r_1>r_2$$
Question 7
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After tracing the path of rays of light through a glass slab for three different angles of incidence, a student measured the corresponding values angle of refraction r and angle of emergence e and recorded them in the table given below:
S.No. $$\angle i$$ $$\angle r$$ $$\angle e$$
I $$30^o$$ $$20^o$$ $$31^o$$
II $$40^o$$ $$25^o$$ $$40^o$$
III $$50^o$$ $$31^o$$ $$49^o$$
The correct observations are:

A
I and II

B
II and III

C
I and III

D
I, II and III

Solution

Condition 1-
Angle of Incidence $$\angle i\ \ \approx$$ Angle of Emergent $$\angle e$$ (Experimental values)
and Angle of Refraction $$\angle r\ < $$ Angle of Incidence $$\angle i$$.

Condition 2-
When Angle of Incidence $$\angle i$$ increase corresponding Angle of Refraction $$\angle r$$ should also increase.
In above Table, all the 3 observations are valid for the conditions.
Question 8
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Above, a light ray is pictured entering a piece of glass from the air.
Which diagram most accurately shows the path the light that passes through the glass will take?

A

B

C

D

E

Solution

When light travels from a rarer medium (i.e air) into the denser medium (i.e glass), it slightly bends towards the normal whereas when it travels from a denser medium into the rarer medium then it bends away from the normal and after it's parallel to incident ray.
Hence option C is correct.
Question 9
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We want a mirror that will make an object look larger. What combination of image and object distances (from the mirror) will accomplish this?

A
Image Distance $$3.0 cm$$, Object Distance $$3.0 cm$$

B
Image Distance $$2.0 cm$$, Object Distance $$3.0 cm$$

C
Image Distance $$3.0 cm$$, Object Distance $$5.0 cm$$

D
Image Distance $$3.0 cm$$, Object Distance $$2.0 cm$$

E
Image Distance $$3.0 cm$$, Object Distance $$10.0 cm$$

Solution

Magnification by a mirror $$m = \dfrac{-v}{u}$$
The image distance $$|u|$$ should be less than object distance $$|v|$$ so that $$m>1$$ i.e. magnified image. Here we don't have to consider the sign of $$u$$ and $$v$$ as we are not concerned about the nature of the imgae. We are interested only in the size of image.

For option D : $$m = \dfrac{- 3.0}{-2.0} = 1.5$$ $$\implies m>1$$
Thus the data given in option D will make a larger image.
Question 10
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The image distance of an object placed 10 cm in front of a thin lens of focal length +5 cm is at :

A
6.5 cm

B
8.0 cm

C
9.5 cm

D
10.0 cm

Solution

Given:
The object distance $$u=-10\ cm$$
The focal length of the lens $$f=+5\ cm$$

From the lens formula,
$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$

$$\Rightarrow \dfrac{1}{5}=\dfrac{1}{v}-\dfrac{1}{-10}$$

$$\Rightarrow \dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{10}=\dfrac{1}{10}$$

$$\therefore v=10\ cm$$

Thus, the distance of the image from the lens is $$v=10cm$$ and it is on the other side of the lens.