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An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length of 2cm.
A. The length of the telescope tube is 20.02m.
B. The magnification is 1000.
C. The image formed is inverted.
D. An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.

Hence, the correct answer is an option (a,b, and c).

Explanation:-

 

  • Use to see heavenly bodies.
  • f_{objective}>f_{eye\; lens} and d_{objective}>d_{eye\; lens}
  • The intermediate image is real, inverted, and small.
  • The final image is virtual, inverte,d and small.

 

  • Magnification:
  • m_{D}=-\frac{f_{0}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right ) and  m_{\infty }=-\frac{f_{0}}{f_{e}}
  • Length :
  • L_{D}=f_{0}+u_{e}=f_{0}+\frac{f_{e}D}{f_{e}+D} and  L_{\infty }=f_{0}+f_{e}

The length of the telescope tube is

f_{0}+f_{e}=20+(0.02)=20.02\; m

Also,

m=\frac{20}{0.02}=1000

Also, the image formed is inverted.

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A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
A. a larger angle to be subtended by the object at the eye and hence viewed in greater detail.
B. the formation of a virtual erect image.
C. increase in the field of view.
D. infinite magnification at the near point.

Hence, the correct answer is option (a,b).

Explanation:-

 A magnifying glass simply contains a convex lens of very small focal length.

For magnification when the final image is formed at D and \infty \left ( \text {i.e}, m_{D}\; \text {and}\; m_{\infty } \right ).

                     m_{D}=\left ( 1+\frac{D}{f} \right )_{max} and m_{\infty }=\left (\frac{D}{f} \right )_{min}

By using a magnifying glass, any object can be brought closer to the eye as compared to the normal near point. The angle formed by the object in the eye is larger and thus, is more detailed. The image formed is virtually erect and is also enlarged.

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Between the primary and secondary rainbows, there is a dark band known as Alexandra’s dark band. This is because

A. light scattered into this region interfere destructively.
B. there is no light scattered into this region.
C. light is absorbed in this region.
D. angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Formed because of light scattered into the region interfere destructively the dark Alexander's band lie between the primary and secondary rainbow; thus, (a, d) are the correct options. The angle of the primary rainbow in the observer's eyes is from 41° to 42°, and in the secondary rainbow, the angle in the observer's eyes lies between 51° to 54° with respect to the incident light ray.

Therefore, the overall angle of scattered rays with reference to incident light of the sun ranges from 42° and 50°.

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A rectangular block of glass ABCD has a refractive index of 1.6. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face AD, the pin shall

A. appears to be near A.
B. appears to be near D.
C. appears to be at the center of AD.
D. not be seen at all.

Hence, the correct answer is option (a,d)

Explanation:-

As given in the figure

 Because till the angle of incidence is less than the critical angle, i.e. incident angle on AD of the ray from the pin, a pin will always seem nearer to point A.

But when the angle of the incident is larger than the critical angle, the light undergoes total internal reflection and is not visible through AD.

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Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because
A. the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
B. the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
C. some of the points of the object far away from the edge may not be visible because of total internal reflection.
D. water in a trough acts as a lens and magnifies the object.

All three (a, b, c) are the correct options.

 When the ray of light passes from water into the air from the pencil it is refracted, the beam bends away from the normal when going from high to low refractive index.

In case of light getting refracted from water surface from the submerged object before reaching to the observer, it bends away from the normal and the angle formed by the from the image of the object is smaller than the real angle formed by the object in the air. This also affects the apparent depth as it becomes nearer to the surface of the water when points are close to the edge as compared to the points far from the edge.

The angle of incident increase while moving right, and it becomes equal to the critical angle. Thus, due to total internal reflection, some point of the object present far from the edge may not be visible.

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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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