Reflection And refraction Test

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Reflection And refraction Test - 3

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Question 1
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An object is initially at a distance of 500 cm from a plane mirror. If the mirror approaches the object at a speed of 2 cm/s, then after 6 seconds, the distance between the object and its image will be

A
476 cm

B
976 cm

C
226 cm

D
150 cm

Solution

In 6 seconds, the distance travelled by the mirror is d = 2 × 6 = 12 cm
Then, the distance between the object and the mirror becomes 500 - 12 = 488 cm
Hence, the distance between the object and its image is 488 × 2 = 976 cm
Question 2
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The image formed by a convex mirror of a real object is larger than the object

A
when u < 2f

B
when u > 2f

C
for all values of u

D
for no value of u

Solution

Image formed by the convex mirror is always diminished in nature. It can not be larger than the size of the object.
Question 3
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A concave mirror of focal length 'f' produces an image 'm' times the size of the object. If the image is real, then the distance of the object from the mirror is

A
(m - 1)f

B

C

D
(m + 1) f

Solution

Using mirror formula, and the magnification, ,
Question 4
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The height of an object is 7 cm and is placed 15 cm from the convex spherical mirror of radius 45 cm. The height of the image formed is

A
14 cm

B
7 cm

C
4.6 cm

D
4.2 cm

Solution

Using mirror formula,
u = -15 cm R = +45 cm

Height of the image is , h is the height of the object.
Question 5
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A ray of light is incident at 50° on the middle of one of the two mirrors arranged at an angle of 60° between them. The ray then touches the second mirror, gets reflected back to the first mirror, making an angle of incidence

A
50°

B
60°

C
70°

D
80°

Solution

Let required angle be θ.

In △ABC,
90° - θ + 140° + 20° = 180°
θ = 70°
Question 6
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A 4.5 cm long needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.

A
6.7 cm; 0.56

B
7.6 cm; 0.23

C
8.3 cm; 0.12

D
9.2 cm; 0.92

Solution

Height of the needle, h₁ = 4.5 cm
Object distance, u = − 12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:

Hence, magnification of the image,
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual and diminished.
Question 7
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A ray of light enters from a rarer medium to a denser medium at an angle of incidence where the reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r', respectively. The critical angle of the denser medium is

A
tan^-1 (sin i)^-1

B
tan^-1 (sin i)

C
sin^-1 (tan i)^-1

D
sin (tan r')

Solution

It is quite clear r + r = 90° or r = 90° - r and i = r
The refractive index is given by = = tan i

Now = tan i ⇒ sin C = (tan i)^-1
So, C = sin^-1 (tan i)^-1
Question 8
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A transparent object is immersed in water. In order that the object becomes invisible, it should

A
behave as a perfect reflector

B
absorb all light falling on it

C
have refractive index one

D
have refractive index exactly matching with that of the surrounding fluid

Solution

From the formula for refraction through a lens, focal length (f) is

where n is refractive index of the material of the lens and R₁, R₂, are the radii of curvature of its surfaces. If the lens is immersed in a liquid whose refractive index is equal to the refractive index of the material of the lens , then
= 1
= (1 – 1) = 0
That is, the focal length of lens becomes infinite. Now, the lens will behave just like a plane transparent plate and will become invisible.
Alternative: When the refractive indices of two mediums are exactly same, then there will not occur refraction and reflection at the common surface of separation due to which the visibility is observed.
Question 9
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The focal length of the objective and the eye piece of a microscope are 4 mm and 25 mm, respectively. If the final image is formed at infinity and the length of the tube is 16 cm, the magnifying power of microscope will be

A
300

B
400

C
500

D
600

Solution

If u₀ , v₀ and the f₀ are the object distance, image distance and the focal length of the objective and if u_e , v_e and the f_e are the object distance, image distance and the focal length of the eye piece.
Length of the tube L = v₀+ u_eand u₀= f₀and L = v₀
Magnifying power of the microscope is
Question 10
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An object is put at a distance of 5 cm from the first focus of a convex lens of focal length 10 cm. If a real image is formed, its distance from the lens will be

A
15 cm

B
20 cm

C
25 cm

D
30 cm

Solution

Here u = -15 cm, f = 10 cm,
Using lens formula,
Question 11
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A point source of light S is placed at a distance L in front of the centre of a plane mirror PQ of width d. It is hung vertically on a wall as shown in the figure. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it. The greatest distance over which he can see the image of the light source in the mirror is

A

B
d

C
2d

D
3d

Solution

Refer to Fig. It is clear that the greatest distance is AB. Now, since PR = RD = L, from triangles
PRT and PBD we have

BD = 2 RT = 2 RS = 2 × = d
and OD = . Therefore, OB = OD + BD
=

AB = 2OB = 2 × = 3d,
which is choice (4).
Question 12
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An air bubble in a glass slab (= 1.5) is 6 cm deep as viewed from one face and 4 cm deep as viewed from other face. The thickness of the glass slab is

A
10 cm

B
6.67 cm

C
15 cm

D
None of these

Solution

Let y₁ is apparent depth as seen from one side and y₂ be the apparent depth as seen from the other side.

Hence real depth d = x₁ + x₂
d = x₁ + x₂ = μ(y₁+ y₂)
d = 1.5(6 + 4) = 15 cm
Question 13
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A thin lens has focal length f and its aperture has diameter d. It forms an image of intensity I. Now, the central part of the aperture up to diameter (d/2) is blocked by an opaque paper. The focal length and the image intensity will change respectively to

A
(f/2) and (I/2)

B
f and (I/4)

C
(3f/4) and (I/2)

D
f and (3I/4)

Solution

As there is no change in the radius of curvature, there will be no change in the focal length.
Intensity of the block, .
Area of central part which is blocked =
Area through which the light will pass is,

Intensity of the final image is .
Question 14
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A microscope has an objective of 5 mm focal length and eye-piece of 30 mm focal length and the distance between them is 150 mm. The magnification is

A
50

B
100

C
200

D
250

Solution

Far point magnifying power of the microscope is given by

Using approximation,
and

Here D = 25 cm or 250 mm
L = 150 mm
Question 15
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In cold countries, the phenomenon of looming (i.e. ship appears in the sky) takes place, because refractive index of air

A
decreases with height

B
increases with height

C
does not change with height

D
becomes infinity at the surface

Solution

In cold countries, the temperature of the land surface is very less. As we move up, the temperature of the air increases. Hence, the refractive index of the air decreases. So, the rays coming from the ship are travelling from the denser medium to the rarer medium and it suffers total internal reflection because of which ship appears in the sky. This phenomenon of optical illusion is called looming.
Question 16
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The path of a ray of light in different media of the refractive indices n₁, n₂, n₃and n₄ is shown in the figure. The velocity of light will be maximum in the medium whose refractive index is

A
n₁

B
n₂

C
n₃

D
n₄

Solution

When the ray of the light is travelling from medium n₂ to n₃, it is moving away from the normal. It means the ray of light is travelling from denser medium (n₂) to rarer medium (n₃). Comparing medium 1 and 3, it is evident that the ray is far away from the normal in the medium 3. So, n₃is the rarest medium and we know that the lesser the refractive index, the greater is the speed of the light.
Question 17
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A thin symmetric double convex lens of power P is cut into three parts A, B and C as shown. The power of

A
A is P, B is

B
A is 2P, B is P

C
A is , B is

D
A is P, B is

Solution

Power of the lens,
(R₁ is the radius of curvature of first refracting surface and R₂ is the radius of curvature of second refracting surface)

For the part A, R₁ = +R, R₂ = -R

The radius of curvature remains the same.
In case of part B, R₁ = +R, R₂ =
Hence,
In case of part C, R₁ =, R₂ = +R. Hence,
Hence, P_A = 2P_B = 2P_C
Question 18
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A student has been given a project to prepare an astronomical telescope. Which two types of lenses does she have to buy?

A
Concave lenses of different focal lengths

B
Convex lenses of same focal length

C
Convex lenses of different focal lengths

D
Concave lenses of same focal lengths

Solution

The astronomical telescope consists of two convex lenses. One is an objective lens of large focal length and an eye-piece of small focal length. Hence, she should select two convex lenses of different focal lengths.
Question 19
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An object is placed at a distance of x₁from the focus on the principal axis of the concave mirror. The image is formed at a distance x₂ from the focus. Then, the focal length of the mirror is

A

B

C

D

Solution

Here u = - (f + x₁), v = - (f + x₂)
Using mirror formula, we have
Question 20
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The image of a candle formed by a convex lens is found to be real and inverted. What change would be found in the nature of this image if the lower half of the lens is covered with black paper?

A
Only the upper half of the candle would be found in the image.

B
There would be no change in the image of any kind.

C
The image would now be real and erect at the same location.

D
The image would now be real and inverted at the same location but its intensity would be reduced to half.

Solution

The position and the size of the image will remain unaffected but the intensity of the image will be reduced to half as the intensity of the image is proportional to the area. As the lower half of the lens is covered with black paper, so the area through which the light is coming will be reduced to half.