Optics

Which one of the following types of radiations has the smallest wave-length ?

[amp_mcq option1=”Microwaves” option2=”Infra-red” option3=”Visible light” option4=”X-rays” correct=”option4″]

The correct answer is (D) X-rays.

Electromagnetic radiation is ordered by wavelength and frequency in the electromagnetic spectrum. The spectrum, from longest wavelength (lowest frequency) to shortest wavelength (highest frequency), includes Radio waves, Microwaves, Infra-red, Visible light, Ultraviolet, X-rays, and Gamma rays. Among the given options, X-rays have the shortest wavelength.

Visible light occupies a small portion of the spectrum, with wavelengths typically ranging from about 400 nm (violet) to 700 nm (red). Infra-red radiation has longer wavelengths than visible light, and Microwaves have even longer wavelengths. X-rays have much shorter wavelengths, typically ranging from 0.01 nm to 10 nm.

Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become

[amp_mcq option1=”double” option2=”half” option3=”four times” option4=”one fourth” correct=”option2″]

The radiation force (F) on a totally reflecting surface is given by F = 2IA/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A₂ = A₁/2. The new radiation force F₂ will be F₂ = 2IA₂/c = 2I(A₁/2)/c = (2IA₁/c) / 2 = F₁/2. Thus, the radiation force on it will become half.

Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.

The radiation pressure is defined as the force per unit area (P = F/A). For a totally reflecting surface, the radiation pressure is P = 2I/c. The force is the product of pressure and area: F = P * A = (2I/c) * A.

A lemon kept in water in a glass tumbler appears to be larger than its actual size. It is because of

[amp_mcq option1=”reflection of light” option2=”scattering of light” option3=”refraction of light” option4=”polarization of light” correct=”option3″]

The correct answer is (C) refraction of light. When light passes from one medium (water) to another (air), it bends. This bending of light is called refraction.

Refraction is the phenomenon of bending of light as it passes from one optical medium to another with different optical density.

The apparent magnification of the lemon under water is due to the refractive indices of water, glass, and air. The light rays from the part of the lemon submerged in water bend away from the normal when they exit the water into the air, making the lemon appear larger and closer than it is.

The correct relation between the radius of curvature R and focal length f of a spherical mirror is

[amp_mcq option1=”R = f” option2=”R = 2f” option3=”R = 3f” option4=”R = 4f” correct=”option2″]

The correct answer is B) R = 2f.

For a spherical mirror (concave or convex) with a small aperture, the focal length (f) is defined as half of its radius of curvature (R). This fundamental relationship is expressed as f = R/2 or R = 2f.

The radius of curvature (R) is the radius of the sphere from which the mirror is a part, and the focal length (f) is the distance from the mirror’s pole to its principal focus (the point where parallel rays converge after reflection for a concave mirror, or appear to diverge from for a convex mirror). This relationship holds true for paraxial rays, i.e., rays close to and parallel to the principal axis.

Radiations coming from the Sun are mostly in the form of

[amp_mcq option1=”light only.” option2=”light and long wavelength infrared.” option3=”light and short wavelength infrared.” option4=”both short and long wavelength infrared.” correct=”option3″]

The correct option is C) light and short wavelength infrared.

The Sun emits electromagnetic radiation across a broad spectrum, with the peak intensity in the visible light region, and significant contributions in the infrared and ultraviolet regions.

The Sun’s surface temperature (around 5500°C or 5778 K) causes it to radiate energy according to Wien’s displacement law and Planck’s law. The peak wavelength falls in the visible spectrum (around 500 nm). The emitted radiation also includes infrared. Due to the high temperature, the infrared radiation from the Sun is predominantly in the short-wavelength range (near-infrared, roughly 0.7 to 5 µm), which is higher energy compared to the long-wavelength infrared emitted by cooler objects like the Earth’s surface (which peaks around 10 µm). Therefore, solar radiation consists mainly of visible light and short wavelength infrared.

Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because

[amp_mcq option1=”the holes through the fibre are extremely fine.” option2=”light rays are absorbed at the entry end and relieved at the exit end of the fibre.” option3=”scattering of light occurs inside the fibre.” option4=”successive total internal reflections occur as a ray moves through the fibre.” correct=”option4″]

The correct answer is D) successive total internal reflections occur as a ray moves through the fibre.

Optical fibres work on the principle of Total Internal Reflection (TIR). The fibre consists of a core with a higher refractive index surrounded by cladding with a lower refractive index. Light entering the core at suitable angles strikes the core-cladding interface at angles greater than the critical angle, causing the light to be reflected back into the core. This process repeats along the length of the fibre, guiding the light even if the fibre is bent.

Option A is irrelevant to the principle of light propagation. Option B is incorrect; light is transmitted, not absorbed and re-emitted in this way. Option C, scattering, generally leads to loss of signal, not efficient transmission through zigzag paths. TIR is the key phenomenon enabling light transmission through optical fibres over long distances with minimal loss.

Which one of the following statements regarding lenses is not correct ?

[amp_mcq option1=”A convex lens produces both real and virtual images.” option2=”A concave lens produces virtual images.” option3=”A convex lens can produce images equal, greater and smaller than the size of the object.” option4=”A concave lens always produces images smaller than the size of the object.” correct=”option4″]

Let’s analyze each statement regarding lenses, assuming standard behaviour with real objects, but considering the possibility of virtual objects as the word “always” appears in option D.
A) A convex lens produces both real and virtual images. This is correct. A convex lens produces real images for objects placed beyond its focal point (F) and virtual images for objects placed between F and the optical center (O).
B) A concave lens produces virtual images. This is correct. A concave lens produces virtual, erect, and diminished images for any real object position. It can also produce real images for virtual objects. The statement “produces virtual images” is true as it is capable of doing so.
C) A convex lens can produce images equal, greater and smaller than the size of the object. This is correct. For real objects, a convex lens produces diminished images (object beyond 2F), same-sized images (object at 2F), and magnified images (object between F and 2F). It also produces a magnified virtual image (object between F and O).
D) A concave lens always produces images smaller than the size of the object. This statement is incorrect. While a concave lens always produces smaller, virtual images for *real* objects, it can produce magnified *real* images for certain *virtual* object positions (specifically, when a virtual object is placed between the focal point F and the optical center O). Therefore, it does not *always* produce images smaller than the object size when considering all possible object types.

– Convex lenses can form both real and virtual, magnified, diminished, or same-sized images depending on the object position.
– Concave lenses always form virtual, erect, and diminished images for real objects.
– The word “always” in statement D makes it universally true or false. Considering virtual objects, a concave lens does not always produce smaller images.

For a concave lens, a virtual object placed between the optical center (O) and the focal point (F) forms a real, inverted, and magnified image beyond F. This contradicts statement D, making it incorrect in the most general sense. If the question were restricted to real objects, all other options would be correct, which suggests D is the intended incorrect statement due to the inclusion of “always”.

An object is placed in front of a convex mirror. Which one of the following statements is correct?

[amp_mcq option1=”It will never form an inverted image.” option2=”The image moves towards the focus when the object moves towards the mirror.” option3=”Depending on the position of the object with respect to the mirror, the image can be inverted and real.” option4=”The size of the image becomes larger than that of the object when the object is placed at a distance equal to half the focal length.” correct=”option1″]

The correct statement for an object placed in front of a convex mirror is A) It will never form an inverted image.

A convex mirror always forms a virtual, erect, and diminished image, regardless of the position of the object (as long as it’s in front of the mirror). Since the image is always erect, it can never be inverted.

– Statement B is incorrect: When the object moves towards a convex mirror, the virtual image also moves towards the mirror (from the focus towards the pole) and increases in size, but it remains virtual and erect.
– Statement C is incorrect: Convex mirrors only form virtual images, never real or inverted images.
– Statement D is incorrect: Convex mirrors always form diminished images (smaller than the object), regardless of the object’s position.

Which one of the following statements is correct for a plane mirror?

[amp_mcq option1=”Its focal length is zero.” option2=”The size of the image of an object placed in front of the mirror is slightly less than that of the object.” option3=”The image is virtual, erect and laterally inverted.” option4=”Its focal length is 200 cm.” correct=”option3″]

The correct statement for a plane mirror is C) The image is virtual, erect and laterally inverted.

A plane mirror forms a virtual image, which means it appears to be behind the mirror and cannot be projected onto a screen. This image is always erect (upright) relative to the object. It is also laterally inverted, meaning the left side of the object appears as the right side of the image, and vice versa.

Other properties of the image formed by a plane mirror include:
– The size of the image is equal to the size of the object.
– The distance of the image behind the mirror is equal to the distance of the object in front of the mirror.
– The focal length of a plane mirror is considered to be infinite.

The refractive indices of two media are denoted by n₁ and n₂ and the velocities of light in these two media are respectively v₁ and v₂. If n₂ / n₁ is 1.5, which one of the following statements is correct?

[amp_mcq option1=”v₁ is 1.5 times v₂.” option2=”v₂ is 1.5 times v₁.” option3=”v₁ is equal to v₂.” option4=”v₁ is 3 times v₂.” correct=”option1″]

The correct statement is that v₁ is 1.5 times v₂.

– The 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 that medium (v). That is, n = c/v.
– For two media, we have n₁ = c/v₁ and n₂ = c/v₂.
– The given relation is n₂ / n₁ = 1.5.
– Substitute the expressions for n₁ and n₂: (c/v₂) / (c/v₁) = 1.5.
– This simplifies to (c/v₂) * (v₁/c) = v₁ / v₂ = 1.5.
– Rearranging the equation, we get v₁ = 1.5 * v₂.

– A higher refractive index means light travels slower in that medium.
– Since n₂/n₁ = 1.5 > 1, it means n₂ > n₁. Therefore, light travels slower in medium 2 (v₂) than in medium 1 (v₁).
– The relationship v₁ = 1.5 v₂ confirms that v₁ is greater than v₂.