Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 57

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Question 1
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A light ray passes from a material of low refractive index to one of high refractive index. Which of the pairs of quantities listed below describes the light ray as it strikes and passes through the interface between the two materials? Match the descriptions below with the statement that best describes the situation.
Statement: A part of the light ray remains inside the low refractive index material.

A
the angle of refraction is larger than the angle of incidence

B
the angle of refraction is equal to the angle of incidence

C
the angle of reflection is larger than the angle of incidence

D
the angle of incidence equals the angle of reflection

E
the angle of reflection is smaller than the angle of incidence

Solution

The part of the light that remains in the medium of low refractive index is actually the reflected ray. We know that in the case of reflection,
$$\text{Angle of incidence} = \text{Angle of reflection}$$

So, the correct option is D.
Question 2
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A ray of light incident on the interface of the medium $$A$$ and $$B$$ and it passes through medium $$B$$ as shown in above figure. Choose the correct option.

A
Medium 1 is more dense than medium 2

B
Medium 1 less dense than medium 2

C
The speed of the light in medium 1 is greater than the speed of the light in medium 2

D
The frequency of the light in medium 1 is greater than the frequency of the light in medium 2

E
The wavelength is the same in medium 1 and medium 2

Solution

When a light ray travels from a rarer medium to denser medium, then the refracted ray bends towards the normal whereas when it travels from denser to rarer medium, then the refracted ray bends away from the normal.
According to the figure, the refracted ray bends away from the normal.
It means it goes from denser to rarer.
Hence medium 1 is more dense than the medium 2.
Question 3
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A beam of light passing from air enters into the glass. Identify which of the following statements is true?

A
The angle of incidence is greater than the angle of refraction in the glass.

B
The angle of incidence is less than the angle of refraction in the glass.

C
The angle of incidence is equal to the angle of refraction in the glass.

D
The frequency of the light decreases.

E
The frequency of the light increases.

Solution

Let the refractive index of air be $$\mu_{a}$$ and that of the glass be $$\mu_{g}$$.
Also let $$i$$ be the angle of incidence and $$r$$ be the angle of refraction.

From Snell's law,
$$\mu_a \sin{i} = \mu_g \sin{}r \Rightarrow \dfrac{\mu_a}{\mu_g}=\dfrac{\sin{r}}{\sin{i}}$$
As $$\mu_g > \mu_a \Rightarrow \sin{i} > \sin{r}\Rightarrow i > r$$
Therefore option A is correct.

The frequency of the light doesn't change during refraction because frequency is a property of the source of light.
Question 4
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Choose the correct option which is true about a bi-concave lens :

A
Its focal length is positive

B
It cannot form real images

C
It cannot form virtual images

D
It can magnify objects

E
None of the above

Solution

A bi-concave lens diverges the light rays after refraction so it is not possible for the rays to meet on other side of lens after refraction because of that we can't obtain the image on other side , it means the image will seem to be formed on the same side of lens ,because light rays seem to be coming out of here ,which cannot be taken on a screen therefore it cannot form a real image.
As seen in diagram it don't form magnified image and it's focal length is negative.
Question 5
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A converging lens has formed an image at a distance of 60 cm from the lens and the height of the image is $$\frac{1}{4}$$ the height of the object. Calculate the focal length of the lens.

A
36 cm

B
45 cm

C
48 cm

D
72 cm

E
80 cm

Solution

It is given that magnification of the lens is $$-\dfrac{1}{4}=\dfrac{v}{u}$$
$$\implies u=-4v=-240cm$$
Using lens equation,
$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$
$$=\dfrac{1}{60}-\dfrac{1}{-240}$$
$$=\dfrac{1}{48}$$
$$\implies f=48cm$$
Question 6
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A ray of light incident on a piece of glass as shown above. The light passes through the glass and emerged into the air. A, B, C, D and E are possible paths of light. Choose the best possible path of light.

A
A

B
B

C
C

D
D

E
E

Solution

According to snell's law of refraction.
At upper interface.
As the ray passes from a less dense medium (air) to a more dense medium (glass), the ray bends toward the normal line.
At lower interface
As the ray exits the glass(denser medium) back into the air(rarer medium), it must bend away from the normal back to the original angle in the air.

hence option D.
Question 7
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The image of a person appears inverted on the far side of an optical instrument . Identify the optical instrument?

A
Concave mirror

B
Convex mirror

C
Plane mirror

D
Concave lens

E
Convex lens

Solution

A mirror cannot form an image on other side of mirror because there is no refraction .
A concave lens always forms an image on the same side of lens because it diverges the rays after refraction so they cannot meet on the other side of lens.
A convex lens can form an image on the other side of lens because it converges the rays after refraction so they can meet on the other side of lens , therefore given lens is convex as image is formed on the other side of the instrument.
Question 8
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In your laboratory, you trace the path of light rays through a glass slab for different values of angle of incidence $$(\angle i)$$ and in each case, measure the values of the corresponding angle of refraction $$(\angle r)$$ and angle of emergence$$(\angle e)$$. Based on your observations, the correct conclusion is

A
$$\angle i$$ is more than $$\angle r$$ , but nearly equal to $$\angle e$$

B
$$\angle i$$ is less than $$\angle r$$ , but nearly equal to $$\angle e$$

C
$$\angle i$$ is more than $$\angle e$$ , but nearly equal to $$\angle r$$

D
$$\angle i$$ is less than $$\angle e$$ , but nearly equal to $$\angle r$$

Solution

Since the light ray enters from a rarer (air) medium to a denser (glass) medium, the ray bends toward normal. As a result, the angle of incidence (angle between the incident ray and normal) $$\angle i$$ is more than the angle of refraction (angle between refracted ray and normal ) $$\angle r$$.
Let $$n_2$$ be the refractive index of denser medium(glass slab).
Let $$n_1$$ be the refractive index of rarer medium(air).
Now using Snell's law, when the light ray goes into glass from the air,
$$\dfrac{n_2}{n_1}=\dfrac{\sin i}{\sin r}$$ ...................$$(1)$$,

When the light ray goes into the air from glass,
$$\dfrac{n_1}{n_2}=\dfrac{\sin r}{\sin e}$$

or $$\dfrac{n_2}{n_1}=\dfrac{\sin e}{\sin r}$$ ........................$$(2)$$

Using $$(1)$$ and $$(2)$$ , we get ,
$$\dfrac{\sin i} {\sin r}=\dfrac{\sin e}{\sin r}$$ ,
$$\sin i = \sin e$$
or $$\angle i=\angle e$$
Thus, $$\angle i$$ is more than $$\angle r $$, but nearly equal to $$\angle e$$.
Question 9
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Study the following four experimental set-ups I, II, III and IV for the experiment, "To trace the path of a ray of light through a rectangular glass slab."
Which of the marked set-ups is likely to give best results (P1 and P2 are the positions of pins fixed on the incident ray) ?

A
I

B
II

C
III

D
IV

Solution

The pins must not be too close to each other, otherwise it would be hard to align the light passing through both of them when seen from the other side of the slab. Hence $$III$$ is rejected.
The incident ray of light must not be normal to the glass slab surface, otherwise it would directly go straight ahead giving no information about the refractive index. Hence $$I$$ is rejected.
Also the angle of incidence must be optimum for observing the shifted path of light after emergence from the other side of slab. A large angle of incidence can give poor results. Hence $$II$$ is rejected.
Correct answer is option D.
Question 10
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On the basis of their experiment, "To trace the path of a ray of light through a rectangular glass slab," students of a class arrived at which one of the following conclusions :

A
Angle of incidence is greater than the angle of emergence,

B
Angle of emergence is smaller than the angle of refraction.

C
Emergent ray is parallel to the refracted ray.

D
Incident ray and emergent ray are parallel to each other.

Solution

The students arrived at the result that angle of incidence and angle of emergence are equal to each other i.e. incident ray and emergent ray are parallel to each other .