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There are certain materials developed in laboratories that have a negative refractive index (Fig. 9.3). A ray incident from the air (medium 1) into such a medium (medium 2) shall follow a path given by

A.

 

 

The answer is the option (a).

Explanation:-

 

  • Option A: The ray is refracted in a direction consistent with a negative refractive index.
  • Option B: This might not match the behavior expected.
  • Option C: A different path that could represent the refraction at the interface.
  • Option D: The light behaves in a very specific way based on the negative refractive index.

In materials with a negative refractive index, light behaves differently than in typical materials with a positive refractive index. The direction of refraction is reversed compared to what you would expect in a normal medium. This means that when a ray of light enters the medium with a negative refractive index, the angle of refraction will be on the opposite side of the normal relative to the angle of incidence.

When a material has a negative refractive index, it follows the Snell's law in an opposite manner. And when an incident ray from one medium (air) falls on them, it bends just the way shown in option (a) which is on the same side as the normal.

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A car is moving at a constant speed of 60\; km\; h^{-1} on a straight road. Looking at the rearview mirror, the driver finds that the car following him is at a distance of 100 m and is approaching at a speed of 5\; km\; h^{-1}. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car every 2s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?

A. The speed of the car in the rear is 65\; km\; h^{-1}.
B. In the side mirror the car in the rear would appear to approach with a speed of 5\; km\; h^{-1} to the driver of the leading car.
C. In the rearview mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
D. In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

The answer is the option (d).

Explanation:-

 When an object is placed in front of a mirror, all the positions of objects in the front of a mirror, image are virtual, erec,t and smaller in size. As the object moves toward the pole, magnification increases and tends to unity at the pole.

An object moving along the principal axis – on differentiating the mirror formula with respect to time we get,

\frac{dv}{dt}=\left ( \frac{v}{u} \right )^{2}\frac{du}{dt}=-\left ( \frac{f}{u-f} \right )^{2}\frac{du}{dt}

Where \frac{dv}{dt} is the velocity of the image moving along the principal axis, and \frac{du}{dt} is the velocity of the object along the principal axis.

The negative sign indicates that the image always moves in the direction opposite to that of the object

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The optical density of turpentine is higher than that of water while its mass density is lower. Fig 9.2. shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Fig 9.2, the path shown is correct?

A. 1

B. 2

C. 3

D. 4

Hence, the correct answer is option (b).

Explanation:-

 

The correct path followed by the ray of light 

 Snell's law describes the relationship between the angle of incidence and the angle of refraction.

\mu _{1}\sin \theta_{1}=\mu _{2}\sin \theta_{2}=\text {Constant}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(1)

Where \mu _{1}\; \text {and}\; \mu _{2} are the refractive indices of the two media

In this diagram, the ray of light is going from an optically rarer medium to turpentine which is an optically denser medium, then the ray bends towards the normal, which means \theta _{1}> \theta _{2}, but when it goes in the opposite direction, i.e.

optically denser medium turpentine to rarer medium water, the ray bends away from the normal.

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The direction of a ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection are shown by four rays marked 1, 2, 3, and 4 (Fig 9.1). Which of the four rays correctly shows the direction of the reflected ray?

Hence, the correct answer is option (b).

Explanation:-

The ray PQ incident to the concave mirror passes through the focus F, after reflection, as shown in the diagram is supposed to be parallel to the principal axis.

 

Important Note:

To find out the exact extended graphical location of the image of any object, one can draw any of the given two rays:

1.  Draw a ray reflected through the focus of the mirror, which is initially parallel to the principal axis (1).

2.  When a ray is drawn through the center of curvature, it reflects back along with itself (3).

3.  Draw a ray that is initiated through the focus; it will be reflected parallel to the principal axis (2).

4.    A ray drawn incident to the pole gets reflected back in a symmetrical manner.

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The phenomena involved in the reflection of radio waves by ionosphere is similar to
A. reflection of light by a plane mirror.
B. total internal reflection of light in air during a mirage.
C. dispersion of light by water molecules during the formation of a rainbow.
D. scattering of light by the particles of air.

 The answer is the option (b)

The phenomena are very similar to total internal reflection of light in the air during a mirage, option (b). The Ionosphere layer of atmosphere around earth is responsible for reflecting the radio waves. And as the angle of incidence is greater than critical, it is very similar to total internal reflection of light in the air during a mirage.

 

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The radius of curvature of the curved surface of a Plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

A. acts as a convex lens only for the objects that lie on its curved side.
B. acts as a concave lens for the objects that lie on its curved side.
C. acts as a convex lens irrespective of the side on which the object lies.
D. acts as a concave lens irrespective of the side on which the object lies.

 The correct answer is option (c).

Explanation:-

Act as a convex lens irrespective of the side on which the object lies. The relation between f,\mu ,R_{1} and R_{2} is known as lens maker's formula and it is

\frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

R_{1}=\infty ,R_{2}=-R

f=\frac{R}{\left ( \mu -1 \right )}          

Here, R=20 cm, \mu =1.5. On substituting the values, we get

f=\frac{R}{\mu -1}=\frac{20}{15-1}=40\; cm

As f>0 means converging nature. Therefore, the lens acts as a convex lens irrespective of the side on which the object lies.

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You are given four sources of light each one providing a light of a single color – red, blue, green, and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
A. The beam of red light would undergo total internal reflection.
B. The beam of red light would bend towards normal while it gets refracted through the second medium.
C. The beam of blue light would undergo total internal reflection.
D. The beam of green light would bend away from the normal as it gets refracted through the second medium.

The answer is the option (c).

Explanation:-

The beam of blue light would undergo total internal reflection, According to the Cauchy relationship, the smaller the wavelength higher the refractive index, and consequently smaller the critical angle.

We know v=f\lambda, the frequency of wave remains unchanged with medium hence v\alpha \lambda

The critical angle,

\sin \; C=\frac{1}{\mu }

Also, the velocity of light

, v\alpha \frac{1}{\mu }

According to VIBGYOR, among all given sources of light, blue light has the smallest wavelength. As \lambda _{blue}<\lambda _{yellow} hence v _{blue}<v _{yellow} it means \mu _{blue}<\mu _{yellow}

It means the critical angle for blue is less than the yellow color, the critical angle is the east which facilitates total internal reflection for the beam of blue light.

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A passenger in an aeroplane shall

A. never see a rainbow.
B. may see a primary and a secondary rainbow as concentric circles.
C. may see a primary and a secondary rainbow as concentric arcs.
D. shall never see a secondary rainbow.

The answer is the option (b)

A passenger has an aeroplane has the possibility to (b) may see a primary and a secondary rainbow as concentric circles because the plane is at an extremely high altitude.

In case of an object moving towards the convex lens with a constant speed from infinity to focus, the image will move slower in the beginning and then faster.

V_{i}=\left ( \frac{f}{f+u} \right )^{2}V_{o}

And if an object approaches the lens, the image moves away from the lens with a non-uniform acceleration.

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An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image

A. moves away from the lens with a uniform speed 5 m/s.
B. moves away from the lens with uniform acceleration.
C. moves away from the lens with a non-uniform acceleration.
D. moves towards the lens with a non-uniform acceleration.

The answer is option (c).

The image will move away from the lens with a non-uniform acceleration. In this case, the objects are moving with a uniform speed of 5 m/s towards a convergent lens from the left; thus, with a non-uniform acceleration, the image moves away from the lens. The image starts at a uniform speed but gets accelerated going from 2F to F; the image goes from 2F to infinity. When the image is at 2F, both the object and image will have the same speed.

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A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

A. blue
B. green
C. violet
D. red

The emerging colour would be (d) red, because of the relation v=f\lambda , which provides the velocity of a wave. During the changing of the mediums, the frequency of the wave remains constant. So, v\; \alpha \lambda or wavelength is directly proportional to speed.

As red colour has the highest wavelength in the spectra, it also has the highest speed according to tot eh relation. Hence, when the light travels in the slab, the first colour to emerge is red.

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