Optics Archives

121. A lady is standing in front of a plane mirror at a distance of 1 m fro

A lady is standing in front of a plane mirror at a distance of 1 m from it. She walks 60 cm towards the mirror. The distance of her image now from herself (ignoring the thickness of the mirror) is

[amp_mcq option1=”40 cm” option2=”60 cm” option3=”80 cm” option4=”120 cm” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2016

Download PDF Attempt Online

In a plane mirror, the image formed is virtual, erect, and located at the same distance behind the mirror as the object is in front of the mirror.
Initially, the lady is at a distance of 1 m (100 cm) from the mirror. Her image is formed 100 cm behind the mirror. The distance between the lady and her image is 100 cm (lady to mirror) + 100 cm (mirror to image) = 200 cm.
The lady walks 60 cm towards the mirror. Her new distance from the mirror is 100 cm – 60 cm = 40 cm.
Her new image is formed 40 cm behind the mirror.
The new distance between the lady and her image is her distance from the mirror + her image’s distance from the mirror = 40 cm + 40 cm = 80 cm.

– Plane mirror property: Object distance from mirror = Image distance from mirror.
– Image is behind the mirror.
– Distance between object and image is (object distance) + (image distance) = 2 * (object distance).

The image in a plane mirror is laterally inverted (left-right reversal). The magnification is +1 (image is same size as object and erect). The concept of image distance behind the mirror being equal to object distance in front is key to solving this problem.

122. A pencil is placed upright at a distance of 10 cm from a convex lens o

A pencil is placed upright at a distance of 10 cm from a convex lens of focal length 15 cm. The nature of the image of the pencil will be

[amp_mcq option1=”real, inverted and magnified” option2=”real, erect and magnified” option3=”virtual, erect and reduced” option4=”virtual, erect and magnified” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2016

Download PDF Attempt Online

A convex lens has a positive focal length, f = 15 cm. The object (pencil) is placed at a distance u = 10 cm from the lens. Since the object is placed in front of the lens, u = -10 cm (using standard sign convention).
For a convex lens, when the object is placed between the optical centre and the principal focus (i.e., 0 < |u| < f, or 0 < 10 cm < 15 cm), the image formed is virtual, erect, and magnified.

– Convex lens image formation depends on object position relative to F (focus) and 2F.
– Object between optical centre and F: Image is virtual, erect, magnified, and on the same side as the object.

Using the lens formula 1/f = 1/v – 1/u:
1/15 = 1/v – 1/(-10)
1/15 = 1/v + 1/10
1/v = 1/15 – 1/10 = (2 – 3)/30 = -1/30
v = -30 cm.
Since v is negative, the image is virtual and formed on the same side as the object.
Magnification m = v/u = -30 / -10 = +3.
Since m is positive, the image is erect. Since |m| > 1, the image is magnified.
Thus, the image is virtual, erect, and magnified.

123. An object is placed at the centre of curvature of a concave mirror of

An object is placed at the centre of curvature of a concave mirror of focal length 16 cm. If the object is shifted by 8 cm towards the focus, the nature of the image would be

[amp_mcq option1=”real and magnified” option2=”virtual and magnified” option3=”real and reduced” option4=”virtual and reduced” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2016

Download PDF Attempt Online

A concave mirror of focal length f = 16 cm has its centre of curvature (C) at a distance 2f = 32 cm from the mirror. Initially, the object is placed at C (u = 32 cm). When the object is at C, the image is formed at C, is real, inverted, and of the same size.
The object is shifted by 8 cm towards the focus (which is at 16 cm). The new object distance is u’ = 32 cm – 8 cm = 24 cm.
The object is now located between the centre of curvature (32 cm) and the focus (16 cm). For a concave mirror, when the object is placed between C and F, the image formed is real, inverted, and magnified.

– Concave mirror image formation depends on object position relative to F (focus) and C (centre of curvature).
– Object at C: Image at C, real, inverted, same size.
– Object between C and F: Image beyond C, real, inverted, magnified.

The mirror formula is 1/f = 1/v + 1/u. Using f = 16 cm and u = 24 cm, we can calculate the image distance v:
1/16 = 1/v + 1/24
1/v = 1/16 – 1/24 = (3 – 2)/48 = 1/48
v = 48 cm.
Since v is positive, the image is real and formed in front of the mirror, beyond C (at 32 cm).
The magnification m = -v/u = -48/24 = -2.
The negative sign indicates an inverted image, and |m| > 1 indicates a magnified image. Thus, the image is real, inverted, and magnified.

124. The Sun is observed to be reddish when it is near the horizon, i.e., i

The Sun is observed to be reddish when it is near the horizon, i.e., in the morning and the evening. This is because

[amp_mcq option1=”red light is least scattered by atmosphere” option2=”red light is most scattered by atmosphere” option3=”it is the colour of the Sun in the morning and evening” option4=”Earth’s atmosphere emits red light” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The scattering of sunlight by particles in the atmosphere is described by Rayleigh scattering. According to Rayleigh scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength (I ∝ 1/λ⁴). This means shorter wavelengths (like blue and violet) are scattered much more effectively than longer wavelengths (like red and orange). When the Sun is near the horizon, sunlight travels through a much greater thickness of the atmosphere. Most of the shorter-wavelength blue light is scattered away from the line of sight, leaving the longer-wavelength red and orange light to reach the observer, making the Sun appear reddish.

The reddish appearance of the Sun at sunrise and sunset is due to the atmospheric scattering of light, specifically the preferential scattering of shorter wavelengths.

The blue colour of the sky during the day is also a result of Rayleigh scattering, where the abundant blue light from the Sun is scattered in all directions across the sky.

125. An object is placed 10 cm in front of a convex lens of focal length 15

An object is placed 10 cm in front of a convex lens of focal length 15 cm. The image produced will be

[amp_mcq option1=”Real and magnified” option2=”Virtual and magnified” option3=”Virtual and reduced in size” option4=”Real and reduced in size” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it’s in front of the lens). Using the lens formula $1/f = 1/v – 1/u$:
$1/15 = 1/v – 1/(-10)$
$1/15 = 1/v + 1/10$
$1/v = 1/15 – 1/10$
$1/v = (2 – 3)/30$
$1/v = -1/30$
$v = -30$ cm
The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image.
The magnification ($m$) is given by $m = v/u$:
$m = (-30) / (-10) = +3$
The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object).
Therefore, the image is virtual and magnified.

Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.

For a convex lens, when the object is placed between the optical center and the focal point (i.e., $u < f$), a virtual, erect, and magnified image is formed on the same side as the object. In this case, $u=10$ cm and $f=15$ cm, so $u < f$, fitting this scenario.

126. A ray of light when refracted suffers change in velocity. In this cont

A ray of light when refracted suffers change in velocity. In this context, which one among the following statements is correct ?

[amp_mcq option1=”Velocity increases as the ray passes from a rarer to a denser medium” option2=”Velocity decreases as the ray passes from a denser to a rarer medium” option3=”Velocity decreases as the ray passes from a rarer to a denser medium” option4=”Change of velocity does not depend on the nature of medium” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

When a ray of light passes from a rarer medium (like air) to a denser medium (like glass or water), its speed decreases. This change in speed is related to the refractive index of the medium; a denser medium has a higher refractive index, causing light to slow down. Conversely, when light passes from a denser medium to a rarer medium, its speed increases.

Understanding how the speed of light changes when it undergoes refraction from one medium to another.

The relationship between the speed of light ($v$), the speed of light in vacuum ($c$), and the refractive index ($n$) of a medium is given by $v = c/n$. A denser medium has a higher refractive index ($n$), resulting in a lower speed of light ($v$). A rarer medium has a lower refractive index, resulting in a higher speed of light.

127. Optical fibres, though bent in any manner, allows light to pass throug

Optical fibres, though bent in any manner, allows light to pass through. What is the inference that one can draw from it ?

[amp_mcq option1=”The concept that light travels in straight path is wrong” option2=”Light can flow through the optical fibres” option3=”Light can travel through the fibres because of their ductility” option4=”Light can travel through the fibres due to multiple total internal reflections” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The correct answer is Light can travel through the fibres due to multiple total internal reflections. Optical fibres work based on the principle of Total Internal Reflection (TIR).

Optical fibres consist of a core (higher refractive index) surrounded by cladding (lower refractive index). Light entering the core at a suitable angle undergoes total internal reflection at the interface between the core and cladding, bouncing back into the core. This process repeats along the length of the fibre, guiding the light even when the fibre is bent, as long as the bend is not too sharp.

Option A is incorrect; light travels in straight lines in a homogeneous medium, but its path changes at interfaces or in non-homogeneous media. Option B is true but doesn’t explain the mechanism allowing travel through bends. Option C is irrelevant; ductility is a mechanical property, not an optical principle.

128. Statement-I: Due to diffused or irregular reflection of light, a close

Statement-I: Due to diffused or irregular reflection of light, a closed room gets light even if no direct sunlight falls inside the room.
Statement-II: Irregular reflection, where the reflected rays are not parallel, does not follow the laws of reflection.

[amp_mcq option1=”Both the statements are individually true and Statement II is the correct explanation of Statement I” option2=”Both the statements are individually true but Statement II is not the correct explanation of Statement I” option3=”Statement I is true but Statement II is false” option4=”Statement I is false but Statement II is true” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

The correct option is C. Statement I is true, but Statement II is false.

Statement I is true. Light entering a room, whether direct sunlight or light from outside reflected by various surfaces, undergoes diffused reflection when it strikes the walls, furniture, and other objects inside the room. Diffused reflection scatters light in multiple directions, illuminating the entire room even areas not directly hit by the initial light rays.
Statement II is false. Irregular reflection (diffused reflection) occurs when light strikes a rough or uneven surface. While the reflected rays scatter in various directions, *each individual ray* still obeys the laws of reflection: the angle of incidence equals the angle of reflection, and the incident ray, the normal to the surface at the point of incidence, and the reflected ray all lie in the same plane. The scattering happens because the normals to the surface at different points are oriented in different directions.

The difference between regular (specular) reflection and irregular (diffused) reflection lies in the nature of the reflecting surface, not in the adherence to the laws of reflection. A smooth surface causes regular reflection (like a mirror), while a rough surface causes diffused reflection (like a wall).

129. Statement-I: Diamond is very bright. Statement-II: Diamond has very lo

Statement-I: Diamond is very bright.
Statement-II: Diamond has very low refractive index.

This question was previously asked in

UPSC NDA-1 – 2015

Download PDF Attempt Online

Statement I is true, but Statement II is false.

Statement I is true because diamond is known for its exceptional brilliance and sparkle, which makes it a highly valued gemstone. Statement II is false; diamond actually has a very *high* refractive index (approximately 2.42), one of the highest among naturally occurring transparent minerals.

The brilliance and fire of a diamond are primarily due to its high refractive index and high dispersion (the ability to split white light into its constituent colours). The high refractive index causes significant bending of light entering the stone and allows for high internal reflection, leading to light bouncing around inside before exiting, creating sparkle. If diamond had a *low* refractive index, light would pass through with less bending and reflection, resulting in significantly less brilliance.

130. The eyepiece of a telescope is a system of two biconvex lenses L1 and

The eyepiece of a telescope is a system of two biconvex lenses L1 and L2, of focal length 30 cm and 20 cm respectively, with a separation of 2 cm between them. Lens L2 faces the eye. This eyepiece is to be replaced by a single bi-convex lens L. Which one among the following is the correct combination of the focal length (F in cm) of the lens L and its distance (δ in cm) from the position of lens L1 ?

[amp_mcq option1=”F = 12·5 and δ = 1·17″ option2=”F = 12·5 and δ = 1·15″ option3=”F = 11·5 and δ = 1·17″ option4=”F = 11·5 and δ = 1·15″ correct=”option1″]

This question was previously asked in

UPSC Geoscientist – 2024

Download PDF Attempt Online

The correct option is A.

The focal length (F) and position (δ) of an equivalent single lens replacing a system of two thin lenses (f₁ and f₂, separated by d) can be calculated using specific formulas related to lens combinations and principal planes. The formula for equivalent focal length is 1/F = 1/f₁ + 1/f₂ – d/(f₁f₂). The distance of the second principal plane (H₂) from the second lens (L₂) is β = -dF/f₁. The distance of the equivalent lens from the first lens (L₁) is typically the distance of the second principal plane from the first lens when light enters from L1 side (δ = d + β).

Given focal lengths f₁ = 30 cm, f₂ = 20 cm, and separation d = 2 cm. L₂ faces the eye, so light comes from L₁ then L₂.

Equivalent focal length F:
1/F = 1/f₁ + 1/f₂ – d/(f₁f₂)
1/F = 1/30 + 1/20 – 2/(30 * 20)
1/F = (2 + 3)/60 – 2/600
1/F = 5/60 – 1/300
1/F = 1/12 – 1/300
1/F = (25 – 1)/300 = 24/300 = 2/25
F = 25/2 = 12.5 cm.

This confirms F = 12.5 cm, eliminating options C and D.

The question asks for the distance (δ) of the equivalent lens L from the position of lens L₁. For a system of lenses, the equivalent single lens producing the same effect as the combination is placed at one of the principal planes. Since L₂ faces the eye, the light exits towards the eye after passing through L₂. The equivalent lens is typically placed at the second principal plane (H₂) of the system, as measured from the lens where light exits (L₂). The position of H₂ is measured relative to L₂.
The distance of the second principal plane H₂ from L₂ is given by β = -d * (F / f₁).
β = -2 cm * (12.5 cm / 30 cm) = -2 * (12.5/30) = -25/30 = -5/6 cm.
The negative sign indicates that H₂ is to the left of L₂ (towards L₁).
The distance of H₂ from L₁ is the distance L₁-L₂ plus the distance L₂-H₂.
Distance L₁-H₂ = d + β = 2 cm + (-5/6 cm) = 2 – 5/6 = (12 – 5)/6 = 7/6 cm.
7/6 cm ≈ 1.1667 cm.
The distance δ from L₁ is approximately 1.17 cm.

This matches option A (F = 12.5 and δ = 1.17).