Physics: Topic-wise Test- 7 - NEET MCQ

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.

The ability of the eye in terms of variation of the focal length to form sharp images on the retina is called accommodation.
 
In simpler terms, the ability of the eye to change the focus from a near distance or a far distance is called accommodation of the eye.
It is possible because of the lens which changes its shape. This makes it possible for the eye to keep focus on an object even when its distance from the eye changes.
 

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.

P_eq=2P_L1+2P_L2+P_M
=(2/fL1)+2(1/f_L2)-(1/f_M)
P_eq=-2(1/12)+(4/45)-(2/-15)
Peq=1/18
-1/f=1/18
F=-18cm
Here the system acts as a concave mirror.

There are three images formed due to refraction at two different lenses:

• One image is created at the top.
• One image results from the refraction through both lenses.
• One image is generated at the bottom.

For the focus of a plano-concave lens, the formula is:

f = R / (1 - μ)

For diagram (ii):

f = R / 2(1 - μ)

For diagram (iii), the same formula applies because it also involves a combination of two plano-concave lenses.

Therefore, the ratio of focal lengths for diagrams (ii) and (iii) is:

1:1

(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

 

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

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).
 

 As per the question, rays are falling on the mirror, which cleared that Regular reflection is taking place.
So, As per the laws of regular reflection.
=> ∠incident= ∠reflection
∠incident= ∠reflection=38A^o
Also we know that normal is at 90°to the mirror.
Now, to get the required angle we need to reduce angle of reflection from normal.
=>90A^o-38A^o=52A^o
so, the correct answer is option B.

If the final rays are converging, we have a real image.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
The correct answer is option B.

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.
 

Let ray strikes the vertical mirror by angle of incidence θ. Then by joining the perpendiculars of two mirrors it is clear from the diagram that angle of incidence at the horizontal mirror is (90−θ).
 
∴ reflected ray from the horizontal mirror makes angle θ with the horizontal mirror also incident ray at vertical mirror makes angle θ with horizontal axis. 
∴ reflected ray is parallel to the original ray.
This is true for two perpendicular mirrors fixed at any positions.

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
 

The number of images between two plane mirror inclined at an angle θ when the object is placed symmetrically between the mirrors is given as
n=(360o​/ θ) −1
Given, 
θ=60o
∴n=(360^o​/60)−1=6−1=5

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.

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.

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.

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.

Every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets.

According to Huygens' principle, a plane light wave propagates though free space at the speed of light, c. The light rays associated with this wave-front propagate in straight-lines. It is also fairly straightforward to account for the laws of reflection and refraction using Huygens' principle.

Here,
Aperture, a = 4 mm = 4 × 10-3 m
Wavelength, λ = 400 nm = 400 × 10-9 m = 4 × 10-7 m 

Ray optics is good approximation upto distances equal to Fresnel's distance (ZF).
Fresnel's distance is given by,

From air to liquid:
1 ->air
2->liquid
 
n₁sinθ₁= n₂sinθ₂
1 x sin35^o=1.48sinθ₂
θ₂=arcsin((1xsin35o)/1.48)=22.8^o

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.