Physics: Topic-wise Test- 7 - NEET MCQ

# Physics: Topic-wise Test- 7 - NEET MCQ

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## 30 Questions MCQ Test NEET Mock Test Series 2025 - Physics: Topic-wise Test- 7

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Physics: Topic-wise Test- 7 - Question 1

### When a plane mirror AB is placed horizontally on level ground at a distance of 60 metres from the foot of a tower, the top of the tower and its image in the mirror subtends, an angle of 90° at B. The height of the tower is:

Detailed Solution for Physics: Topic-wise Test- 7 - Question 1

The angle subtended at the eye by the top of the tower with respect to the ground is 45 degrees. This happens when the height of the tower is equal to the distance from the eye. The eye is assumed to be very near to the mirror. (Refer below figure).
Angle subtended by AB = 90 degrees/2 = 45 degrees.
Therefore, tanθ=h/60​. That is, tan45=h/60​.So, h=60m.
Hence, the height of the tower is 60 m.

Physics: Topic-wise Test- 7 - Question 2

### A unnumbered wall clock shows time 04 : 25 : 37, where 1st term represents hours, 2nd represents minutes & the last term represents seconds, What time will its image in a plane mirror show.

Detailed Solution for Physics: Topic-wise Test- 7 - Question 2

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Physics: Topic-wise Test- 7 - Question 3

### Two plane mirrors of length L are separated by distance L and a man M2 is standing at distance L from the connecting line of mirrors as shown in figure. A man M1 is walking in a straight line at distance 2L parallel to mirrors at speed u, then man M2 at O will be able to see image of M1 for total time :

Detailed Solution for Physics: Topic-wise Test- 7 - Question 3

The light from man M1​ hits the mirror and then reaches the man M2​. The positions of the moving man need to be found when the light from him hits the corners of the mirror to reach the standing man.
From geometry,
AO′=L/2​+2(L/2​)
=3L/2
​Similarly, BO′=(3L/2​)+2(3L/2​)
=9L/2
Hence AB=(9L/2)-(3L/2)=3/2
Time taken to cover this distance=3L/u​
Similarly time taken to cover CD=3L/u​
Thus total time he is visible is 6L/u​

Physics: Topic-wise Test- 7 - Question 4

Which of the following cannot form real image of a real object ?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 4

A concave mirror can form both real and virtual images depending upon the distance between object and mirror.

A convex mirror can only form virtual, erect, and diminished images; therefore, it is true that it can never form a real image of a real object.

-Virtual image cannot be formed from a virtual object.
- When an object is placed between pole and focus, the image formed by the concave mirror is magnified, virtual and erect.
-When an object is placed beyond the centre of curvature , image of it is formed between centre of curvature and focus which is diminished,real and inverted,so real image of real object can be formed by concave mirror.
-When the object is virtual, the image formed is real for the concave surface as shown in figure.

*Multiple options can be correct
Physics: Topic-wise Test- 7 - Question 5

The radius of curvature of the left & right surface of the concave lens are 10 cm & 15 cm respectively. The radius of curvature of the mirror is 15 cm.

Detailed Solution for Physics: Topic-wise Test- 7 - Question 5

Peq=2PL1+2PL2+PM
=(2/fL1)+2(1/fL2)-(1/fM)
Peq=-2(1/12)+(4/45)-(2/-15)
Peq=1/18
-1/f=1/18
F=-18cm
Here the system acts as a concave mirror.

*Multiple options can be correct
Physics: Topic-wise Test- 7 - Question 6

If a symmetrical biconcave thin lens is cut into two identical halves. They are placed in different ways as shown :

Detailed Solution for Physics: Topic-wise Test- 7 - Question 6

There form 3 images due to refraction on two different lenses 3 times one upper ,1 refraction from both lenses, 1 at the bottom.
now focus of Plano concave =f=R/(1−μ)
hence focus for diagram ii> f=R/2(1−μ)
and same focus for diagram iii>because again there is combination of two two Plano concave lens
so, the ratio of focus is 1:1

*Multiple options can be correct
Physics: Topic-wise Test- 7 - Question 7

By properly combining two prisms made of different materials, it is possible to

Detailed Solution for Physics: Topic-wise Test- 7 - Question 7

We can either have deviation or not and same for dispersion as only μ is different and angle of prism is of our choice
option A,B,C are correct.

Physics: Topic-wise Test- 7 - Question 8

A transparent sphere of radius R and refractive index m. An object O is placed at a distance x from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface.

If x = 2R, then the value of m is

Detailed Solution for Physics: Topic-wise Test- 7 - Question 8

(μ/v)- (μ/u) = (μ2- μ1)/R
Or, (μ/2R) + (1/2R) = μ-1/R
Or, (μ/2) + (1/2) = (μ-1)/1
Or, (Μ+1)/2= μ-1
Or, μ+1=2μ-2
μ=3

Physics: Topic-wise Test- 7 - Question 9

A transparent sphere of radius R and refractive index m. An object O is placed at a distance x from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface.

If x = ∞, then the value of m is

Detailed Solution for Physics: Topic-wise Test- 7 - Question 9

From the symmetry of the fig.ray inside the sphere is parallel to the principal axis. Taking refraction at A
2/v)−(μ1/u))=(μ2−μ1)/R
(μ/∞)−(1/−R)=(μ−1)/R
(1/R)=(μ−1)/R⇒μ−1=1
μ=2

Physics: Topic-wise Test- 7 - Question 10

Column -II shows the optical phenomenon that can be associated with optical components given in column-I. Note that column-I may have more than one matching options in column-II.

Physics: Topic-wise Test- 7 - Question 11

Statement-I : If a source of light is placed in front of rough wall its image is not seen.

Statement-II : The wall does not reflect light.

Physics: Topic-wise Test- 7 - Question 12

Statement-I : As the distance x of a parallel ray from axis increases, focal length decreases

Statement-II : As x increases, the distance from pole to the point of intersection of reflected ray with principal axis decreases

Physics: Topic-wise Test- 7 - Question 13

Statement-I : When an object dipped in a liquid is viewed normally, the distance between the image and the object is indepedent of the height of the liquid above the object.

Statement-II : The normal shift is independent of the location of the slab between the object and the observer.

Detailed Solution for Physics: Topic-wise Test- 7 - Question 13

a) At first the height of the liquid level is obviously dependent because the normal shift will happen according to the depth of the liquid.
Drawing a ray diagram keeping a coin at very deep and another close to the surface.We will notice that the earlier one will have a noticeable change of position but the later one will only change a bit.
b) The position of the slab is independent because wherever we keep it the incident ray & the emergent ray at last will be parallel (same medium keeping in mind).

Physics: Topic-wise Test- 7 - Question 14

Statement-I : When two plane mirrors are kept perpendicular to each other as shown ( O si the point object), 3 image will be formed.

Statement-II : In case of multiple reflection, image of one surface can act as an object for the next surface.

Physics: Topic-wise Test- 7 - Question 15

An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will be

Detailed Solution for Physics: Topic-wise Test- 7 - Question 15

Given that,
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100−(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5×6
D=30cm
At this instant,
Distance between mirror and object
D=100−30
D=70cm
Now, the distance between image and object is
D=70−(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.

Physics: Topic-wise Test- 7 - Question 16

A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 16

The number of images formed when two mirrors are inclined to each other is given by :
n=(360/θ -1)
here θ= 90°[since walls are perpendicular]
so, number of images=n=360/90-1
=4-1
=3
These 3 images are formed by a combination of two adjacent walls with the object itself acts as objects for the ceiling mirror. So totally 4 images are formed.
Hence total number of images formed are 4+3= 7

Physics: Topic-wise Test- 7 - Question 17

In image formation from spherical mirrors, only paraxial rays are considered because they

Detailed Solution for Physics: Topic-wise Test- 7 - Question 17

Paraxial rays are ones which originate from a point source and make a small angle with the principal axis. After reflection, they give a point image, either real or virtual.
But if rays from point source that are far from the mirror axis, gives a blurred image, an effect called spherical aberration.

Physics: Topic-wise Test- 7 - Question 18

A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle 1° on the earth. Then the diameter of the image is (in cm)

Detailed Solution for Physics: Topic-wise Test- 7 - Question 18

Given: A concave mirror of radius of curvature 20cm forms image of the sun. The diameter of the sun subtends an angle 1o on the earth.

To find the diameter of the image

Solution:

As sun is more far from the earth,

Then we take u = infinite

Focal length,

From the mirror formula

d = π/18

Physics: Topic-wise Test- 7 - Question 19

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment

Detailed Solution for Physics: Topic-wise Test- 7 - Question 19

Distance between the slits, d=0.28×10−3 m

Distance between the slits and the screen, D=1.4m

Distance between the central fringe and the fourth (n=4) fringe,

u=1.2×10−2 m

In case of a constructive interference, we have the relation for the distance between the two fringes as:

u=nλD/d

⇒λ=ud/nD=6×10−7m=600nm

Physics: Topic-wise Test- 7 - Question 20

Diffraction also refers to

Detailed Solution for Physics: Topic-wise Test- 7 - Question 20

Diffraction is defined as the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the waveforms produced.

Physics: Topic-wise Test- 7 - Question 21

Shape of the wave front of light emerging out of a convex lens when a point source is placed at its focus.

Detailed Solution for Physics: Topic-wise Test- 7 - Question 21

Plane. The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

Physics: Topic-wise Test- 7 - Question 22

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

Detailed Solution for Physics: Topic-wise Test- 7 - Question 22

Physics: Topic-wise Test- 7 - Question 23

A beam of light has a wavelength of 650 nm in vacuum. What is the wavelength of these waves in the liquid whose index of refraction at this wavelength is 1.47?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 23

We know that,
Wavelength of light in a material:
λ= λ0/n
where, λ=wavelength of material, λ0=wavelength of light in vacuum,
n=index of refraction of material.
So, 650x10-9m/1.47=442.177nm.

Physics: Topic-wise Test- 7 - Question 24

wave front is

Physics: Topic-wise Test- 7 - Question 25

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 25

Physics: Topic-wise Test- 7 - Question 26

A parallel beam of light in air makes an angle of 47.5 with the surface of a glass plate having a refractive index of 1.66. What is the angle between the reflected part of the beam and the surface of the glass?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 26

The angle of incidence = 90 - 47.5 = 42.5
Now by snells’ laws we get
sin r = sin 42.5 / 1.66
Thus we get r = 24.01
Thus the angle with surface = 90 -24.01 = 65.99

Physics: Topic-wise Test- 7 - Question 27

Shape of the wave front of portion of the wave front of light from a distant star intercepted by the Earth

Detailed Solution for Physics: Topic-wise Test- 7 - Question 27

Plane (a small area on the surface of a large sphere is nearly planar). The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Physics: Topic-wise Test- 7 - Question 28

Which of the following phenomenon cannot be explained by diffraction?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 28

The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin film interference.

Physics: Topic-wise Test- 7 - Question 29

The critical angle for total internal reflection at a liquid–air interface is 42.5 If a ray of light traveling in air has an angle of incidence at the interface of 35 what angle does the refracted ray in the liquid make with the normal?

Detailed Solution for Physics: Topic-wise Test- 7 - Question 29

From air to liquid:
1 ->air
2->liquid

n1sinθ1= n2sinθ2
1 x sin35o=1.48sinθ2
θ2=arcsin((1xsin35o)/1.48)=22.8o

Physics: Topic-wise Test- 7 - Question 30

According to Huygens construction relation between old and new wave fronts is

Detailed Solution for Physics: Topic-wise Test- 7 - Question 30

The Huygens-Fresnel principle states that every point on a wavefront is a source of wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. The new wavefront is a line tangent to all of the wavelets.
The correct answer is option B.

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