Optics Archives

91. The human eye is like a camera that has a lens with :

The human eye is like a camera that has a lens with :

[amp_mcq option1=”fixed focal length and fixed aperture size.” option2=”variable focal length and fixed aperture size.” option3=”fixed focal length and variable aperture size.” option4=”variable focal length and variable aperture size.” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2024

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The human eye functions much like a camera, possessing features analogous to both a lens and an aperture, both of which can adjust.

The lens of the human eye (specifically the crystalline lens and the cornea) changes its shape through a process called accommodation. This change in shape alters the focal length of the lens, allowing the eye to focus on objects at different distances, similar to a zoom lens in a camera. The iris controls the size of the pupil, which acts as the aperture. By changing the pupil size, the eye regulates the amount of light entering the retina, similar to adjusting the aperture size (f-number) on a camera.

Therefore, the human eye has both variable focal length (through accommodation by the lens) and variable aperture size (controlled by the iris regulating the pupil). Options A, B, and C describe cameras with limitations that the human eye overcomes through its dynamic adjustments.

92. Which one of the following telescopes contains only mirrors ?

Which one of the following telescopes contains only mirrors ?

[amp_mcq option1=”Galilean telescope” option2=”Keplerian telescope” option3=”Newtonian telescope” option4=”Schmidt telescope” correct=”option3″]

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UPSC NDA-1 – 2023

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A Newtonian telescope is a type of reflecting telescope that uses a concave primary mirror to collect and focus light, and a flat secondary mirror placed in the light path to redirect the light to an eyepiece positioned at the side of the telescope tube. The primary and secondary mirrors are the key optical components responsible for forming the image. It uses only mirrors for the main light path and image formation.

– Telescopes are classified as refractors (using lenses), reflectors (using mirrors), or catadioptric (using both lenses and mirrors).
– Galilean and Keplerian telescopes are types of refracting telescopes, using lenses.
– A Newtonian telescope is a type of reflecting telescope, using mirrors (a primary parabolic mirror and a secondary flat mirror).
– A Schmidt telescope (e.g., Schmidt-Cassegrain) uses both a corrector plate (lens) and mirrors.
– The question asks for a telescope containing *only* mirrors in its main optical path for image formation.

Reflecting telescopes, like the Newtonian, have advantages over refracting telescopes, such as the absence of chromatic aberration (color fringing) which occurs in lenses, and they can be made much larger since mirrors are supported from the back.

93. In the dispersion of white light by a common glass prism, which one am

In the dispersion of white light by a common glass prism, which one among the following is correct ?

[amp_mcq option1=”Red light deviates the most because red light has highest speed in prism” option2=”Blue light deviates the most because blue light has highest speed in prism” option3=”Red light deviates the most because red light has lowest speed in prism” option4=”Blue light deviates the most because blue light has lowest speed in prism” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2023

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When white light passes through a prism, it disperses into its constituent colors (VIBGYOR) because the refractive index of the prism material is different for different wavelengths of light. This phenomenon is called dispersion. The amount of deviation experienced by a color depends on the refractive index for that color; greater refractive index leads to greater deviation. For most transparent materials like glass, the refractive index is higher for shorter wavelengths (blue, violet) and lower for longer wavelengths (red). Since the speed of light in a medium is given by v = c/n (where c is the speed of light in vacuum and n is the refractive index), a higher refractive index means a lower speed. Therefore, blue light (shorter wavelength, higher refractive index) deviates the most and has the lowest speed in the prism, while red light (longer wavelength, lower refractive index) deviates the least and has the highest speed.

– Dispersion occurs because the refractive index (n) of the medium depends on the wavelength (λ).
– In most transparent materials (normal dispersion), refractive index decreases as wavelength increases (n_violet > n_blue > … > n_red).
– Deviation (δ) in a prism depends on the refractive index (for a small angle prism, δ ≈ (n-1)A). Higher n means greater deviation.
– Speed of light in the medium (v) is inversely proportional to the refractive index (v = c/n). Higher n means lower speed.
– Thus, blue light has a higher refractive index, deviates the most, and has the lowest speed in the prism.

The order of colors in the dispersed spectrum is typically remembered by the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Violet light deviates the most, followed by Indigo, Blue, etc., with Red deviating the least.

94. If the absolute refractive indices of glass and water are 3/2 and 4/3

If the absolute refractive indices of glass and water are 3/2 and 4/3 respectively, what will be the ratio of velocity of light in glass and water ?

[amp_mcq option1=”3 : 4″ option2=”4 : 3″ option3=”8 : 7″ option4=”8 : 9″ correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2017

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The absolute refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v): n = c / v.
Given the absolute refractive index of glass, n_g = 3/2.
The speed of light in glass, v_g = c / n_g = c / (3/2) = 2c/3.
Given the absolute refractive index of water, n_w = 4/3.
The speed of light in water, v_w = c / n_w = c / (4/3) = 3c/4.
The ratio of the velocity of light in glass and water is v_g / v_w.
v_g / v_w = (2c/3) / (3c/4) = (2c/3) * (4/3c) = (2/3) * (4/3) = 8/9.
The ratio is 8 : 9.

This question tests the relationship between refractive index and the speed of light in a medium. The formula n = c/v is key.

Refractive index is a measure of how much the speed of light is reduced when it passes through a medium compared to its speed in a vacuum. A higher refractive index means light travels slower in that medium.

95. Shown in the figure are two plane mirrors XY and YZ (XY ⊥ YZ) joined a

Shown in the figure are two plane mirrors XY and YZ (XY ⊥ YZ) joined at their edge. Also shown is a light ray falling on one of the mirrors and reflected back parallel to its original path as a result of this arrangement. The two mirrors are now rotated by an angle θ to their new position X’Y’Z’, as shown. As a result the new reflected ray is at an angle α from the original reflected ray. Then :

[amp_mcq option1=”α = 0″ option2=”α = θ” option3=”α = 2θ” option4=”α = 4θ” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2023

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For a ray of light reflecting successively from two plane mirrors placed perpendicular to each other, the final reflected ray is always anti-parallel to the incident ray, regardless of the initial angle of incidence.

This property is characteristic of a corner reflector made of two perpendicular mirrors. The total deviation after two reflections is 180 degrees. When the mirrors are rotated together by an angle θ, the angle between them remains 90 degrees. Therefore, the property holds true, and the final reflected ray remains anti-parallel to the incident ray.

Since the original reflected ray is anti-parallel to the incident ray and the new reflected ray is also anti-parallel to the *same* incident ray, the two reflected rays are parallel to each other and in the same direction (anti-parallel to the incident ray). The angle between them, α, is 0.

96. A rectangle ABCD is kept in front of a concave mirror of focal length

A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A’B’C’D’ in front of the mirror. What is the ratio of B’C’ to A’D’ ?

[amp_mcq option1=”1″ option2=”2″ option3=”1/2″ option4=”2/3″ correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2023

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The ratio of B’C’ to A’D’ is 1/2.

For a concave mirror, the magnification (m) is given by m = -v/u, where v is the image distance and u is the object distance. The size of the image perpendicular to the axis (like B’C’ and A’D’) is related to the object size (BC and AD) by the magnification: Image size = |m| * Object size. Since ABCD is a rectangle, AD = BC.

Using the mirror formula 1/f = 1/u + 1/v:
For point A at u_A = -2f (assuming standard sign convention where f is negative for concave mirror, but distances are given as 2f and 3f. Let’s assume f > 0 is magnitude, and u is negative): u_A = -2f. 1/f = 1/(-2f) + 1/v_A => 1/v_A = 1/f + 1/(2f) = 3/(2f) => v_A = 2f/3. No, if A is at 2f from the mirror, and f is the focal length, then 2f is the radius of curvature C. Object at C forms image at C. So u_A = -2f, then v_A = -2f. Magnification m_A = -v_A/u_A = -(-2f)/(-2f) = -1. A’D’ = |m_A| * AD = AD.
For point B at u_B = -3f: 1/f = 1/(-3f) + 1/v_B => 1/v_B = 1/f + 1/(3f) = 4/(3f) => v_B = 3f/4. Wait, the standard formula for concave mirror with positive f is 1/f = 1/u + 1/v. Let’s use this and assume distances are positive. u_A = 2f, u_B = 3f, f=f. 1/v_A = 1/f – 1/u_A = 1/f – 1/(2f) = (2-1)/(2f) = 1/(2f) => v_A = 2f. Magnification m_A = -v_A/u_A = -(2f)/(2f) = -1. A’D’ = |-1| * AD = AD.
1/v_B = 1/f – 1/u_B = 1/f – 1/(3f) = (3-1)/(3f) = 2/(3f) => v_B = 3f/2. Magnification m_B = -v_B/u_B = -(3f/2)/(3f) = -1/2. B’C’ = |-1/2| * BC = 1/2 * BC.
Since AD = BC, the ratio B’C’ / A’D’ = (1/2 * BC) / AD = 1/2 * (BC/AD) = 1/2 * 1 = 1/2.

97. What is the magnification produced by a concave lens of focal length 1

What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?

[amp_mcq option1=”2.0″ option2=”1.0″ option3=”0.5″ option4=”0.33″ correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2022

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The correct option is C.

For a lens, the lens formula is $1/f = 1/v – 1/u$ and magnification is $m = v/u$. For a concave lens, the focal length $f$ is negative, and it forms virtual, upright, and diminished images, typically on the same side as the object (so image distance $v$ is negative).

Given focal length $f = -10$ cm (concave lens). The image is formed at a distance of 5 cm from the lens. Since it’s a concave lens, the image is virtual and formed on the same side as the object, so image distance $v = -5$ cm.
Using the lens formula: $1/f = 1/v – 1/u$
$1/(-10) = 1/(-5) – 1/u$
$-1/10 = -1/5 – 1/u$
$1/u = -1/5 + 1/10$
$1/u = -2/10 + 1/10$
$1/u = -1/10$
$u = -10$ cm (object distance is 10 cm in front of the lens).
Magnification $m = v/u = (-5 \text{ cm}) / (-10 \text{ cm}) = 0.5$.
The magnification is 0.5, indicating a diminished image.

98. The twinkling of a star is due to the atmospheric

The twinkling of a star is due to the atmospheric

[amp_mcq option1=”diffraction of starlight” option2=”reflection of starlight” option3=”refraction of starlight” option4=”dispersion of starlight” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2022

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The twinkling of stars is caused by the atmospheric refraction of starlight. As starlight passes through the Earth’s atmosphere, which has layers of varying density and temperature, it undergoes continuous refraction. The turbulent movement of air causes fluctuations in the refractive index of the atmosphere along the path of light. This leads to variations in the apparent position and intensity (brightness) of the star as perceived by an observer, resulting in the twinkling effect.

Atmospheric refraction causes starlight to bend as it passes through layers of air with different optical densities, leading to fluctuations in apparent brightness and position.

Planets are closer to Earth than stars and appear as extended sources rather than point sources. The light from planets undergoes similar refraction but the effect is averaged out over their larger apparent disk, which is why planets typically do not twinkle.

99. Mirage is an illustration of

Mirage is an illustration of

[amp_mcq option1=”only dispersion of light.” option2=”only reflection of light.” option3=”only total internal reflection of light.” option4=”both refraction and total internal reflection of light.” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2021

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The correct answer is (D) both refraction and total internal reflection of light.

Mirages are atmospheric optical illusions caused by the bending of light (refraction) as it passes through layers of air with different temperatures and thus different refractive indices. Near hot surfaces (like roads or sand in deserts), the air is hotter and less dense than the air above, meaning its refractive index is lower. Light from the sky or distant objects travels downwards into these layers of increasingly rarer (hotter) air. It bends away from the normal. If the angle of incidence becomes greater than the critical angle at the boundary between two air layers, total internal reflection occurs. This reflected light travels upwards to the observer’s eye, creating the illusion of an image displaced from the actual object’s position.

Refraction is the initial bending of light through the varying density/temperature layers. Total internal reflection occurs when light traveling from a denser medium to a rarer medium hits the boundary at an angle greater than the critical angle, and all light is reflected back into the denser medium. Both processes are necessary for the formation of most common mirages (like inferior mirages).

100. If a ray of light enters from a rarer medium to a denser medium at zer

If a ray of light enters from a rarer medium to a denser medium at zero angle of incidence, it would

[amp_mcq option1=”reflect back.” option2=”go straight.” option3=”turn towards right.” option4=”bend at 45°.” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2021

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The correct answer is (B) go straight.

When a ray of light is incident normally (at zero angle of incidence) on the boundary between two different media, it passes from one medium to the other without changing its direction, regardless of whether it is moving from a rarer to a denser medium or vice versa. The angle of incidence is the angle between the incident ray and the normal to the surface. If this angle is zero, the angle of refraction (the angle between the refracted ray and the normal) will also be zero, as predicted by Snell’s Law (n1 sinθ1 = n2 sinθ2).

Snell’s Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. If θ1 = 0, then sinθ1 = 0. Thus, n2 sinθ2 = 0. Since n2 (refractive index of the second medium) is not zero, sinθ2 must be zero, which means θ2 = 0. Therefore, the refracted ray also lies along the normal, meaning it goes straight. Some partial reflection may occur at the boundary, but the transmitted light goes straight.