Optics Archives

141. Which one of the following statements is correct regarding the travel

Which one of the following statements is correct regarding the travel of a light beam from a rare to a dense medium?

[amp_mcq option1=”A light beam travelling from a rare medium to a dense medium slows down and bends towards the normal.” option2=”A light beam travelling from a rare medium to a dense medium speeds up and bends towards the normal.” option3=”A light beam travelling from a rare medium to a dense medium slows down and bends away from the normal.” option4=”A light beam travelling from a rare medium to a dense medium speeds up and bends away from the normal.” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2023

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When a light beam travels from a rarer medium (lower refractive index, e.g., air) to a denser medium (higher refractive index, e.g., glass or water), its speed decreases. This change in speed causes the light ray to bend unless it is incident normally to the surface.

According to Snell’s Law of refraction (n₁ sin θ₁ = n₂ sin θ₂), where n₁ and n₂ are the refractive indices of the first and second media, and θ₁ and θ₂ are the angles of incidence and refraction respectively (measured from the normal). If light goes from a rarer medium (n₁) to a denser medium (n₂), then n₂ > n₁. For the equation to hold, sin θ₂ must be less than sin θ₁, which means θ₂ < θ₁ (for angles in the first quadrant). A smaller angle of refraction (θ₂) compared to the angle of incidence (θ₁) means the light ray bends towards the normal.

Conversely, when light travels from a denser medium to a rarer medium, its speed increases, and it bends away from the normal. Refractive index is defined as the ratio of the speed of light in vacuum to the speed of light in the medium (n = c/v). A denser medium has a higher refractive index and thus a lower speed of light.

142. A concave mirror of radius of curvature 50 cm is used to form an image

A concave mirror of radius of curvature 50 cm is used to form an image of an object kept at a distance of 25 cm from the mirror on its principal axis. What will be the position of the image from the mirror ?

[amp_mcq option1=”At infinity” option2=”At 50 cm” option3=”At 25 cm” option4=”At 75 cm” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2023

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The correct answer is A) At infinity.

The radius of curvature (R) of a concave mirror is 50 cm. The focal length (f) of a spherical mirror is half the radius of curvature, so f = R/2 = 50 cm / 2 = 25 cm. For a concave mirror, the focal length is considered negative in standard sign conventions when light comes from the left, so f = -25 cm. The object is kept at a distance of 25 cm from the mirror, which means the object distance (u) is -25 cm (object is real, placed on the left).
We use the mirror formula: 1/f = 1/v + 1/u, where v is the image distance.
Substituting the values: 1/(-25) = 1/v + 1/(-25)
-1/25 = 1/v – 1/25
1/v = -1/25 + 1/25
1/v = 0
This implies v = infinity.

When an object is placed at the focal point of a concave mirror, the rays of light from the object become parallel after reflection. Parallel rays meet at infinity, hence forming a real, inverted, and infinitely large image at infinity. This principle is used in devices like searchlights and headlights, where a light source is placed at the focus to produce a parallel beam of light.

143. Which one of the following statements about the aperture of a convex l

Which one of the following statements about the aperture of a convex lens is correct ?

[amp_mcq option1=”It is equal to its radius of curvature.” option2=”It is equal to its focal length.” option3=”It is independent of its radius of curvature.” option4=”It is equal to half of its focal length.” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2022

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The aperture of a convex lens is the effective diameter of the circular outline of the lens from which light passes. It is related to the physical size of the lens and how much light it can collect. The aperture is independent of the lens’s radius of curvature and focal length. While these properties determine how the lens bends light, the aperture determines the brightness of the image and affects phenomena like diffraction and depth of field.

Aperture is a measure of the size of the opening that limits the amount of light passing through a lens. A larger aperture allows more light to enter, resulting in a brighter image and typically a shallower depth of field.

Focal length depends on the radii of curvature of the lens surfaces and the refractive index of the lens material (Lensmaker’s formula). Aperture is a geometric property related to the physical size of the lens or the diaphragm opening within it, and it can be adjusted in optical systems like cameras. It is not determined by or equal to the radius of curvature or focal length.

144. If an object is placed at the focus of a convex lens, its image is

If an object is placed at the focus of a convex lens, its image is

[amp_mcq option1=”at the focus on the same side.” option2=”at the focus on the opposite side.” option3=”coincident with the lens.” option4=”at infinity.” correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2022

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When an object is placed at the principal focus of a convex lens, the rays of light from the object become parallel after refraction through the lens. Parallel rays are considered to meet at infinity. Therefore, the image is formed at infinity.

A convex lens converges parallel rays of light to its principal focus and makes rays originating from its principal focus parallel. This is one of the standard ray diagrams for convex lenses.

The image formed at infinity is typically considered real, inverted, and highly magnified. This principle is used in devices like telescopes (for distant objects) and in projectors to obtain a magnified image at a large distance.

145. Refraction of light, as it enters from one transparent medium to anoth

Refraction of light, as it enters from one transparent medium to another, is due to

[amp_mcq option1=”change in temperature of the media” option2=”change in the amplitude of light” option3=”change in speed of light” option4=”internal property of light” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2021

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Refraction of light occurs when light passes from one transparent medium to another because the speed of light changes as it enters a different medium. This change in speed causes the light ray to bend at the interface between the two media.

The change in the speed of light as it moves from one medium to another is the fundamental reason for refraction. The amount of bending is related to the refractive index of the media, which is defined based on the speed of light in the medium.

When light enters a denser medium (higher refractive index), its speed decreases, and it bends towards the normal. When it enters a rarer medium (lower refractive index), its speed increases, and it bends away from the normal. Option A is incorrect; while temperature can slightly affect the refractive index, it’s not the primary cause of the bending itself. Option B refers to amplitude, related to intensity, not direction. Option D is too vague; the speed change is the specific physical property responsible.

146. A ray of light travelling from a rarer medium to a denser medium

A ray of light travelling from a rarer medium to a denser medium

[amp_mcq option1=”slows down and bends away from the normal.” option2=”slows down and bends towards the normal.” option3=”speeds up and bends away from the normal.” option4=”speeds up and bends towards the normal.” correct=”option2″]

This question was previously asked in

UPSC CDS-2 – 2021

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When a ray of light travels from a rarer medium (like air) to a denser medium (like glass or water), its speed decreases. Due to the change in speed, the light ray bends towards the normal to the surface at the point of incidence.

Refraction is the bending of light as it passes from one medium to another.

Conversely, when light travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. This phenomenon is governed by Snell’s Law.

147. Where should an object be placed in front of a convex lens to get a re

Where should an object be placed in front of a convex lens to get a real and enlarged image of the object ?

[amp_mcq option1=”At twice the focal length” option2=”At infinity” option3=”Between the principal focus and twice the focal length” option4=”Beyond twice the focal length” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2021

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When an object is placed between the principal focus (F) and twice the focal length (2F) of a convex lens, a real, inverted, and enlarged image is formed beyond 2F on the other side of the lens.

The position and nature of the image formed by a convex lens depend on the object’s position relative to the focal point and 2F.

Placing the object at twice the focal length (2F) produces a real, inverted image of the same size at 2F. Placing the object at infinity produces a real, inverted, diminished image at the focal point (F). Placing the object beyond twice the focal length (beyond 2F) produces a real, inverted, diminished image between F and 2F.

148. Which one of the following combinations of source and screen would pro

Which one of the following combinations of source and screen would produce sharpest shadow of an opaque object?

[amp_mcq option1=”A point source and an opaque screen” option2=”An extended source and an opaque screen” option3=”A point source and a transparent screen” option4=”An extended source and a transparent screen” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2020

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The sharpness of a shadow depends on the size and type of the light source and the properties of the screen. A point source of light produces a sharp shadow with a well-defined edge, called the umbra. An extended source of light produces a shadow with a dark central region (umbra) and a partially illuminated region around it (penumbra), resulting in a blurry edge. An opaque object is necessary to block the light and form a shadow. The screen needs to be opaque to display the shadow, as light would pass through a transparent screen. Therefore, the combination of a point source and an opaque screen produces the sharpest shadow of an opaque object.

A point source of light creates a sharp shadow (umbra) without a penumbra, unlike an extended source which creates both umbra and penumbra.

The size and distance of the object and the screen also affect the shadow’s size and shape, but the sharpness is primarily determined by the nature of the light source (point vs. extended) and the screen (opaque vs. transparent). An opaque screen is essential to intercept the light and make the shadow visible.

149. When a light ray enters into glass medium from water at an angle of in

When a light ray enters into glass medium from water at an angle of incidence 0°, what would be the angle of refraction?

[amp_mcq option1=”90°” option2=”45°” option3=”0°” option4=”The ray will not enter at all” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2020

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When a light ray enters a medium from another medium at an angle of incidence of 0°, this means the ray is incident normally (perpendicularly) to the boundary surface between the two media. According to Snell’s Law of Refraction, n₁ sin(i) = n₂ sin(r), where n₁ and n₂ are the refractive indices of the two media, i is the angle of incidence, and r is the angle of refraction. If i = 0°, then sin(i) = sin(0°) = 0. The equation becomes n₁ * 0 = n₂ sin(r), which simplifies to 0 = n₂ sin(r). Since the refractive index of glass (n₂) is not zero, sin(r) must be zero. The angle whose sine is zero is 0°. Therefore, the angle of refraction (r) is 0°. This means the ray passes straight through the boundary without bending.

A light ray incident normally on a boundary between two media passes undeviated, meaning the angle of refraction is also 0°.

Refraction, or the bending of light, occurs when light changes speed as it passes from one medium to another at an angle. When the incidence is normal (angle of incidence = 0), the change in speed still occurs, but there is no change in direction.

150. A luminous object is placed at a distance of 40 cm from a converging l

A luminous object is placed at a distance of 40 cm from a converging lens of focal length 25 cm. The image obtained in the screen is

[amp_mcq option1=”erect and magnified” option2=”erect and smaller” option3=”inverted and magnified” option4=”inverted and smaller” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2020

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For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1/f = 1/v – 1/u to find the image distance (v).
1/25 = 1/v – 1/(-40)
1/25 = 1/v + 1/40
1/v = 1/25 – 1/40 = (8 – 5) / 200 = 3 / 200
v = 200/3 cm ≈ +66.7 cm.
Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted.
The magnification is given by m = v/u = (200/3) / (-40) = -200 / 120 = -5/3 ≈ -1.67.
Since |m| = 5/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified.

For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).

If the object were placed beyond 2f, the image would be real, inverted, and diminished. If the object were at 2f, the image would be real, inverted, and of the same size at 2f on the other side. If the object were between the optical center and f, the image would be virtual, erect, and magnified, formed on the same side as the object.