Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 68

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Weekly Quiz Competition

Question 1
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Which of the following formula is wrong in accordance with Snell's law ?

A
$$ \dfrac{n_1}{\sin i} = \dfrac{n_2}{\sin r} $$

B
$$ \dfrac{n_{2}}{n_1} = \dfrac{\sin i}{\sin r} $$

C
$$n_{1} \sin i = n_{2} \sin r$$

D
$$\dfrac{1}{n_2} \sin i = \dfrac{1}{n_1} \sin r $$

Solution

Snell's Law of refraction states that the product of sine of angle made by the ray in a medium and the refractive index of the medium is constant.
i.e., $$n_{1} \times \sin i = $$ Constant
$$\therefore n_{1} \sin i = n_{2} \sin r $$

$$\Rightarrow \boxed{\dfrac{n_1}{n_2} = \dfrac{\sin r}{\sin i}} $$

Correct Option is A.
Question 2
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Refractive indices of water and glass are 4/3 and 3/2 respectively. A ray of light travelling in water is incident on the water-glass interface at $$30^0$$. Calculate the sine of angle of refraction.

A
0.460

B
0.585

C
0.444

D
0.623

Solution

Angle of incidence $$i = 30^o$$

Refractive index of water, $$n_w = \dfrac{4}{3}$$

Refractive index of water, $$n_g = \dfrac{3}{2}$$

Let the angle of refraction be $$r$$.

Using Snell's law of refraction :

$$n_w \sin i = n_g \sin r$$

$$\therefore$$ $$\dfrac{4}{3}\times \sin 30^o = \dfrac{3}{2} \sin r$$

Or $$\dfrac{4}{3}\times 0.5 = \dfrac{3}{2} \sin r$$ $$(\because \sin 30^o = 0.5)$$

$$\implies \sin r = 0.444$$
Question 3
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An object is placed on the principal axis of a concave mirror at a distance of 60 cm. If the focal length of the concave mirror is 40 cm then determine the magnification of the obtained image.

A
2

B
-2

C
2.5

D
-2.5

Solution

Focal length $${f}=-40cm$$ (-ve sign appears because the mirror is concave)
Object distance $${u}=-60cm$$
Magnification$$(m)=?$$
Using mirror formula
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{-1}{40}=\dfrac{1}{v}+\dfrac{-1}{60}$$
$$\dfrac{1}{v}=\dfrac{-3+2}{120}$$
$$v=-120cm$$
Magnification $$m=\dfrac{-v}{u}=\dfrac{120}{-60}=-2$$ (The minus sign signifies that the image is inverted)
Question 4
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You are given water, mustard oil, glycerine, and kerosene. In which of these media a ray of light is incident obliquely at same angle would bend the most ?

A
Kerosene

B
Water

C
Mustard oil

D
Glycerine

Solution

Refractive index for water, mustard oil, glycerine and kerosene are:
$$\mu_w = 1.33$$
$$\mu_m = 1.47$$
$$\mu_g = 1.473$$
$$\mu_k = 1.44 $$
Since refractive index of glycerine is highest, ray bends the most in case of glycerine.
Question 5
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A pin that is 2 cm long is placed at a distance of 16 cm from a convex lens. Assuming it to be perpendicular to the principal axis, find the position of the image if the focal length of the lens is 12 cm.

A
48 cm

B
24 cm

C
12 cm

D
None of these

Solution

Here $$u=-16 cm$$ and $$f=+12 cm$$

We have $$\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {1}{f}$$

or, $$\dfrac {1}{v}=\dfrac {1}{u}+\dfrac {1}{f}$$ $$=\dfrac {1}{-16 }+\dfrac {1}{12 }=\dfrac {1}{48}$$

or, $$v=+48\ cm$$

The image is formed 48 cm from the lens on the side of the transmitted rays. The image is, therefore, real.
Question 6
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A rod of length 10 cm lies along the principal axis of concave mirror of focal length 10 cm in such a way thatits end closer to the pole is 20 cm away from the mirror. The length of the image is then -

A
15 cm

B
2.5 cm

C
5 cm

D
10 cm

Solution

Given for the concave mirror :
Focal Length, $$f=-10cm$$
Length of the rod, $$L_{AB}=10cm$$
Closer end distance of rod, $$u_A=-20cm$$
Far end distance of rod, $$u_B=u_A+L_{AB}$$
$$u_B=-20-10=-30cm$$
By using mirror formula for end $$A$$:

$$\dfrac{1} {V_{A}} + \dfrac{1} {-20} = \dfrac{1} {-10}$$

$$V_{A} = -20 cm$$ [Image position of the end A]

$$\dfrac{1} {V_{B}} + \dfrac{1} {-30} = \dfrac{1} {-10}$$

$$\dfrac{1} {V_{B}} = \dfrac{1} {15}$$

$$v_{B} = -15 cm$$
Length of the image can be given as:
$$L_{A'B'} = \left |V_{A} \right | - \left |V_{B} \right | = 20 - 15 = 5 cm$$
Question 7
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A concave mirror forms the real image of an object which is magnified 4 times. The object is moved 3 cm away, the magnification of the image is 3 times. What is the focal length of the mirror?

A
3 cm

B
4 cm

C
12 cm

D
36 cm

Solution

For mirror $$u=\frac {f(m-1)}{m}$$
In first case, $$u=\frac {f(-4-1)}{-4}$$
In the second case, $$u+3=\frac {f(-3-1)}{-3}$$
On solving, we get $$f=36 cm$$
Question 8
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A convex mirror has a focal length of 20 cm. A real object is placed at a distance of 20 cm from the pole of the mirror and in front of the mirror. The mirror produces an image at :

A
Infinity

B
20 cm

C
40 cm

D
10 cm

Solution
Question 9
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The magnitude of focal length of a concave mirror is $$f$$. An object is placed at a distance $$x$$ from the focus and forms a real image. The magnification is:

A
$$-\dfrac{f}{x}$$

B
$$-\dfrac{x}{f}$$

C
$$-\dfrac{f}{f-x}$$

D
$$\dfrac{x}{x-f}$$

Solution

The magnification of a concave mirror is given by:
$$m = -\dfrac{f}{f-u} $$
Now, for the given case, $$u$$ is the object distance which is equal to: $$u = f + x$$
Substituting this value in the expression for magnification, we get:
$$m = -\dfrac{f}{f - (f + x)} $$
This, gives us the value of $$m$$ to be $$ \dfrac{-f}{x} $$
Question 10
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Directions For Questions

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under:
Position of candle $$=$$ 12.0 cm
Position of convex lens $$=$$ 50.0 cm
Position of the screen $$=$$ 88.0 cm

...view full instructions

What is the focal length of the convex lens ?

A
20 cm

B
40 cm

C
19 cm

D
12 cm

Solution