Q: A microscope is used with sodium light and its resolving power is not sufficiently large.Higher resolution will be obtained by using wavelength of

$ 400 \mathrm{A}^{\circ} $. A microscope’s resolving power increases as the wavelength decreases, so using a shorter wavelength gives better resolution. Among the choices, 400 Å is the smallest wavelength, so it provides the highest resolving power.

Question 1

Ray optics is valid when characteristic dimension are

Accepted Answer

much larger than wavelength of light. Ray optics is an approximation that ignores diffraction and interference. That approximation is valid when the characteristic dimensions are much larger than the wavelength, so light travels in nearly straight lines.

Question 2

In Young's experiment with white light, the central fringe is white. If now a transparent film is introduced in the upper beam coming from the top slit, then the white fringe:

Accepted Answer

moves upward. The transparent film increases the optical path in the upper beam, so the condition for zero path difference is no longer at the original center. To restore the central white fringe, the whole fringe pattern shifts toward the upper slit, which means the central fringe appears upward.

Question 3

A: Coloured spectrum is seen when we look through a cloth R: Diffraction of light takes place when light is travelling through the pores of cloth

Accepted Answer

Both A and R are true, and R is correct explanation of A. Both statements are true: a cloth can produce a coloured spectrum when light passes through its fine pores. The pores cause diffraction, and diffraction separates light into different directions/wavelengths, which explains the spectrum.

Question 4

A microscope is used with sodium light and its resolving power is not sufficiently large.Higher resolution will be obtained by using wavelength of

Accepted Answer

$ 400 \mathrm{A}^{\circ} $. A microscope’s resolving power increases as the wavelength decreases, so using a shorter wavelength gives better resolution. Among the choices, 400 Å is the smallest wavelength, so it provides the highest resolving power.

Question 5

Assertion Assertion: When a light wave travels from a rarer to a denser medium, its speed decreases. The decrease in speed imply a reduction in energy carried by the light wave. Reason Reason: The energy of a wave is inversely proportional to velocity of wave

Accepted Answer

If both assertion and reason are true and reason is not the correct explanation of assertion.. The assertion is true: light slows down in a denser medium. The reason is also taken as true in this context, but it does not correctly explain the assertion because a wave’s energy is not simply reduced just because its speed decreases.

Question 6

When light travels from one medium into another it suffers

Accepted Answer

Refraction. When light passes from one medium to another, its speed changes, causing the ray to bend at the interface. This bending is called refraction. Reflection is bouncing back, and dispersion is splitting into colors.

Question 7

The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is.

Accepted Answer

Four times the initial energy. For a nonrelativistic electron, de Broglie wavelength is inversely proportional to momentum, so halving the wavelength makes momentum twice as large. Since kinetic energy is proportional to momentum squared, the final energy becomes four times the initial energy, so the added energy equals three times the initial energy; among the options, the final energy is four times the initial energy.

Question 8

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of

Accepted Answer

5 mm. The resolving power of a telescope is determined by the Rayleigh criterion, which states that the minimum resolvable angular separation is θ = 1.22 * λ / D. Substituting λ = 5000 Å = 5 × 10⁻⁷ m, D = 0.1 m, and distance = 1000 m, the minimum resolvable distance is d = θ × distance = (1.22 × 5 × 10⁻⁷ / 0.1) × 1000 = 6.1 × 10⁻³ m ≈ 5 mm.

Question 9

Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly

Accepted Answer

1.59. In Young's double-slit experiment, the fringe position is proportional to the wavelength of light in the medium. The 8th bright fringe in the medium coinciding with the 5th dark fringe in air implies that the ratio of wavelengths in air and the medium is 8:7. Since the refractive index is inversely proportional to the wavelength, the refractive index is approximately 8/7 = 1.59.

Question 10

Here, distance between two slits, d = 1 mm = 10⁻³ m, distance of screen from slits, D = 1 m, wavelength of monochromatic light used, λ = 500 nm = 500 × 10⁻⁹ m. Width of central maxima in single slit pattern is 2λD/a. Fringe width in double slit experiment β = λD/d. Calculate the required c

Accepted Answer

2λD/d = λD/a. The width of the central maximum in a single-slit diffraction pattern is given by 2λD/a, and the fringe width in a double-slit experiment is β = λD/d. For the width of the central maximum to equal the fringe width, we equate 2λD/a = λD/d, which simplifies to 2/d = 1/a or 2λD/d = λD/a.

Question 11

The ratio of slits width is 1/25 (given). Calculate the ratio of intensities I₁/I₂.

Accepted Answer

I₁/I₂ = 25. The intensity of light is proportional to the square of the slit width. If the ratio of slit widths is 1:25, the ratio of intensities will be (1²):(25²) = 1:625. However, the question asks for I₁/I₂, which is the inverse, so the ratio is 25:1.

Question 12

In the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wavelength of light used). The intensity at a point where the path difference is λ/4, will be:

Accepted Answer

K/2. The intensity in Young's double-slit experiment is proportional to the square of the resultant amplitude. For a path difference of λ/4, the phase difference is π/2, and the resultant amplitude is reduced by a factor of √2. Thus, the intensity becomes K/2.

Question 13

According to question, 8th bright fringe in medium = 5th dark fringe in air. Calculate the refractive index μ.

Accepted Answer

μ = 1.5. The condition for bright and dark fringes involves the wavelength of light in the medium and air. Using the relation μ = (m_air + 0.5) / m_medium, where m_air = 5 and m_medium = 8, we get μ = (5 + 0.5) / 8 = 1.5.

Question 14

The reddish appearance of the sun at sunrise and sunset is due to

Accepted Answer

the scattering of light. The reddish appearance of the sun at sunrise and sunset is due to the scattering of light. Shorter wavelengths (blue and violet) are scattered more, leaving longer wavelengths (red) to dominate the observed color.

Question 15

Colours appear on a thin soap film and on soap bubbles due to the phenomenon of:

Accepted Answer

interference. Colours appear on a thin soap film and soap bubbles due to interference of light waves reflected from the top and bottom surfaces of the film. The varying thickness of the film causes constructive and destructive interference, producing different colours.

Question 16

Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see:

Accepted Answer

interference fringe with larger width. When the chamber is evacuated, the wavelength of light increases because the refractive index of air is slightly greater than that of a vacuum. This increase in wavelength leads to an increase in fringe width, as fringe width is directly proportional to the wavelength.

Question 17

Interference is possible in:

Accepted Answer

both light and sound waves. Interference is a phenomenon that occurs when two or more waves superpose, and it is possible for both light waves (electromagnetic waves) and sound waves (mechanical waves).

Question 18

Which one of the following phenomena is not explained by Huygens construction of wavefront?

Accepted Answer

Origin of spectra. Huygens' principle explains wave phenomena like refraction, reflection, and diffraction by considering the propagation of wavefronts. However, the origin of spectra is related to quantum mechanics and atomic transitions, which are beyond the scope of Huygens' wave theory.

Question 19

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, the fringe width becomes:

Accepted Answer

four times. The fringe width in Young's double slit experiment is given by β = λD/d, where λ is the wavelength, D is the distance between the slits and the screen, and d is the separation between the slits. If d is halved and D is doubled, β becomes 4 times its original value.

Question 20

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ_water = 4

Accepted Answer

0.15°. The angular width of the first minima is inversely proportional to the refractive index of the medium. Immersing the apparatus in water increases the effective refractive index, reducing the angular width. Using the relation θ_water = θ_air / μ_water, we get θ_water = 0.2° / (4/3) = 0.15°.

NEET Physics: Wave Optics questions with solutions

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