151. Which of the following lenses will bend the light rays through largest

Which of the following lenses will bend the light rays through largest angle?

[amp_mcq option1=”Lens with power +2·0 D” option2=”Lens with power +2·5 D” option3=”Lens with power –1·5 D” option4=”Lens with power –2·0 D” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2020
The power of a lens is a measure of its ability to converge or diverge light rays. It is defined as the reciprocal of the focal length in meters (P = 1/f). A lens with a higher absolute value of power bends light rays through a larger angle. Comparing the absolute values of the given powers: |+2.0 D| = 2.0 D, |+2.5 D| = 2.5 D, |-1.5 D| = 1.5 D, |-2.0 D| = 2.0 D. The largest absolute power is 2.5 D, corresponding to the lens with power +2.5 D. Therefore, this lens will bend the light rays through the largest angle.
The power of a lens (absolute value) indicates the extent to which it bends light; higher power means greater bending.
The sign of the power indicates the type of lens: positive power for a converging (convex) lens and negative power for a diverging (concave) lens. Converging lenses bring parallel rays together, while diverging lenses spread them out. However, the magnitude of bending is determined by the absolute value of the power.

152. If the speed of light in air is represented by c and the speed in a me

If the speed of light in air is represented by c and the speed in a medium is v, then the refractive index of the medium can be calculated using the formula

[amp_mcq option1=”v / c” option2=”c / v” option3=”c / (2. v)” option4=”(c – v) / c” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2020
The refractive index of the medium can be calculated using the formula c / v.
The refractive index (n) of a medium is a dimensionless quantity that describes how fast light travels through the medium. It is defined as the ratio of the speed of light in vacuum (c₀) to the speed of light in the medium (v). Air is often used as an approximation for vacuum, so if the speed of light in air is represented by c and the speed in the medium is v, the refractive index is given by n = c / v.
The refractive index is always greater than or equal to 1 (n ≥ 1), with n=1 for vacuum (or approximately for air). A higher refractive index means light travels slower in the medium and bends more when entering from a medium with a lower refractive index.

153. Soap solution used for cleaning purpose appears cloudy. This is due to

Soap solution used for cleaning purpose appears cloudy. This is due to the fact that soap micelles can

[amp_mcq option1=”refract light” option2=”scatter light” option3=”diffract light” option4=”polarize light” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2019
Soap solutions, when used for cleaning, form micelles. Micelles are aggregates of soap molecules that are dispersed throughout the water. These aggregates are large enough (typically in the colloidal size range) to scatter light that passes through the solution. This scattering of light, known as the Tyndall effect, makes the solution appear cloudy or opaque.
The cloudy appearance of soap solution is due to the scattering of light by soap micelles (Tyndall effect).
Scattering of light is a characteristic property of colloidal solutions, suspensions, and emulsions, where dispersed particles are larger than the wavelength of light, unlike true solutions where solute particles are too small to scatter visible light.

154. Magnification is

Magnification is

[amp_mcq option1=”actual size of specimen / observed size” option2=”observed size of specimen / actual size” option3=”actual size of specimen – observed size” option4=”actual size of specimen × observed size” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2019
Magnification is a measure of the ability to enlarge the image of an object. It is calculated as the ratio of the size of the image as observed (e.g., through a microscope or on a screen) to the actual size of the specimen being viewed.
Magnification = Observed size of specimen / Actual size of specimen.
Resolution is another important concept related to microscopy, referring to the ability to distinguish between two closely spaced objects as separate entities. Magnification enlarges the image, while resolution allows for clarity and detail.

155. When a convex lens produces a real image of an object, the minimum dis

When a convex lens produces a real image of an object, the minimum distance between the object and image is equal to

[amp_mcq option1=”the focal length of the convex lens” option2=”twice the focal length of the convex lens” option3=”four times the focal length of the convex lens” option4=”one half of the focal length of the convex lens” correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2018
The minimum distance between an object and its real image formed by a convex lens is four times the focal length (4f).
For a convex lens, a real image is formed when the object is placed outside the focal point (object distance |u| > f). The image formed is real and inverted. The lens formula is 1/v – 1/u = 1/f. Using distances as positive values, 1/v + 1/|u| = 1/f. Let D be the distance between the object and the image, D = |u| + v. To minimize D, we can express v in terms of |u| and f: 1/v = 1/f – 1/|u| = (|u|-f)/(f|u|), so v = f|u|/(|u|-f). Thus, D = |u| + f|u|/(|u|-f) = (|u|(|u|-f) + f|u|)/(|u|-f) = (|u|² – f|u| + f|u|)/(|u|-f) = |u|²/(|u|-f). Let x = |u|-f, so |u| = x+f. D = (x+f)²/x = (x² + 2xf + f²)/x = x + 2f + f²/x. For a real image, |u| must be greater than f, so x > 0. By AM-GM inequality, x + f²/x ≥ 2√(x * f²/x) = 2f. The minimum value occurs when x = f²/x, i.e., x² = f², so x = f (since x>0). This means |u|-f = f, so |u| = 2f. When |u|=2f, v = f(2f)/(2f-f) = 2f²/f = 2f. The minimum distance D = |u| + v = 2f + 2f = 4f. This happens when the object is placed at 2f, forming a real image at 2f.
If the object is placed closer than the focal length (|u| < f), a virtual image is formed. As the object approaches the focal point from outside (from |u| > f), the real image moves further away from the lens towards infinity. As the object moves away from the lens, the real image moves towards the focal point. The minimum separation between object and a real image occurs specifically when the object is placed at 2f.

156. The visible portion of the electromagnetic spectrum is

The visible portion of the electromagnetic spectrum is

[amp_mcq option1=”infrared” option2=”radiowave” option3=”microwave” option4=”light” correct=”option4″]

This question was previously asked in
UPSC CDS-2 – 2018
The correct answer is D) light.
The electromagnetic spectrum encompasses a wide range of wavelengths, from radio waves to gamma rays. The visible portion of this spectrum, which is perceived by the human eye, is commonly referred to as visible light. This range includes the colors of the rainbow, from red (longer wavelengths) to violet (shorter wavelengths).
Infrared radiation has longer wavelengths than visible light and is associated with heat. Radiowaves and microwaves have much longer wavelengths and are used for communication and heating (like in microwave ovens), respectively. None of these are visible to the human eye. The visible spectrum lies between approximately 380 and 750 nanometers.

157. Infrared, visible and ultraviolet radiations/ light have different pro

Infrared, visible and ultraviolet radiations/ light have different properties. Which one of the following statements related to these radiations/light is not correct?

[amp_mcq option1=”The wavelength of infrared is more than that of ultraviolet radiation.” option2=”The wavelength of ultraviolet is smaller than that of visible light.” option3=”The photon energy of visible light is more than that of infrared light.” option4=”The photon energy of ultraviolet is lesser than that of visible light.” correct=”option4″]

This question was previously asked in
UPSC CDS-2 – 2017
The statement “The photon energy of ultraviolet is lesser than that of visible light” is not correct.
Electromagnetic radiations are ordered by their frequency or wavelength. In order of increasing energy (and frequency, and decreasing wavelength), the spectrum goes from Infrared (IR) to Visible light to Ultraviolet (UV). Photon energy is directly proportional to frequency (E = hf) and inversely proportional to wavelength (E = hc/λ). Therefore, UV radiation has higher frequency, higher photon energy, and shorter wavelength compared to visible light and infrared radiation.
Statement A is correct because IR has a longer wavelength than UV. Statement B is correct because UV has a shorter wavelength than visible light. Statement C is correct because visible light has higher energy than IR. Statement D claims UV energy is lesser than visible light energy, which is false; UV energy is higher than visible light energy.

158. A ray of light is incident on a plane mirror at an angle of 40° with r

A ray of light is incident on a plane mirror at an angle of 40° with respect to normal. When it gets reflected from the mirror, it undergoes a deviation of

[amp_mcq option1=”40°” option2=”100°” option3=”90°” option4=”80°” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2017
When a ray of light is incident on a plane mirror at an angle of 40° with respect to normal, it undergoes a deviation of 80°.
According to the law of reflection, the angle of incidence (i) is equal to the angle of reflection (r), and both are measured with respect to the normal to the mirror surface. Given i = 40°, then r = 40°. Deviation is the angle between the direction of the incident ray and the direction of the reflected ray. If the incident ray were not reflected, it would continue in a straight line. The reflection changes its direction.
The angle between the incident ray and the normal is 40°. The angle between the reflected ray and the normal is 40° on the opposite side of the normal. The total angle between the incident ray and the reflected ray is the sum of the angle between the incident ray and the normal and the angle between the normal and the reflected ray in the plane of reflection, which is 40° + 40° = 80°. This angle represents the deviation of the ray from its original path. Alternatively, the angle between the incident ray and the mirror surface is θ = 90° – i = 90° – 40° = 50°. The angle of deviation (δ) for reflection from a plane mirror is also given by δ = 180° – 2θ = 180° – 2(50°) = 180° – 100° = 80°.

159. Light travels in a straight line (rectilinear propagation of light). T

Light travels in a straight line (rectilinear propagation of light). This statement does hold if the medium of travel for light is

[amp_mcq option1=”of variable refractive index” option2=”made up of slabs of different refractive indices” option3=”homogeneous and transparent” option4=”inhomogeneous and transparent” correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2016
Light travels in a straight line in a homogeneous and transparent medium. A homogeneous medium is one where the refractive index is constant throughout. Transparency allows light to pass through.
Rectilinear propagation of light, or the principle that light travels in straight lines, is a fundamental concept in ray optics and holds true when the medium is uniform and has a constant optical density (homogeneous) and allows light to pass through (transparent).
If the medium has a variable refractive index (A) or is made of different slabs with different refractive indices (B), light will bend (refract) as it moves through the medium or crosses boundaries between regions. An inhomogeneous medium (D) also implies a varying refractive index, causing light to not travel in a straight line.

160. In total internal reflection, the light travels from

In total internal reflection, the light travels from

[amp_mcq option1=”rarer to denser medium and it occurs with no loss of intensity” option2=”denser to rarer medium and it occurs with no loss of intensity” option3=”rarer to denser medium and it occurs with loss of intensity” option4=”denser to rarer medium and it occurs with loss of intensity” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2016
Total internal reflection (TIR) is a phenomenon that occurs when light rays travel from a denser medium to a rarer medium and strike the boundary at an angle greater than the critical angle. At this angle, all the light is reflected back into the denser medium. This process is considered lossless because ideally, no light is transmitted or absorbed at the boundary.
For total internal reflection to occur, two conditions must be met: light must travel from a denser medium to a rarer medium, and the angle of incidence must exceed the critical angle. It is characterized by minimal to no loss of intensity upon reflection.
When light travels from a rarer to a denser medium, reflection and refraction occur, but total internal reflection is not possible. Reflection at an interface typically involves some loss of intensity due to absorption or incomplete reflection. Total internal reflection is used in technologies like fiber optics and prisms due to its high efficiency in reflecting light.