Wave Optics
33 Board Physics previous year questions on Wave Optics — options free on every question; 3 include the answer & explanation free, the rest unlock with PYQ Pass.
A ring of radius 3 cm has a soap film which is getting evaporated. Light of wavelength \(\lambda=580 \mathrm{~nm}\) gives minimum transmission every 12 s . Find the rate of evaporation. (Refractive index=1.45)(Shift - II Memory Based)
\(\begin{aligned}& 15 \pi \times 10^{-12} \mathrm{~m}^3 / \mathrm{s}\end{aligned}\)
To find the rate of evaporation of the soap film, we need to determine the rate at which its thickness decreases over time.
Step 1: Understanding the Given Data- Wavelength of light: \(\lambda =580\) = \(580\times 1{0}^{−9}\)
- Refractive index of soap film: \(\mu =1.45\)
- Minimum transmission occurs every \(T=12\)
- Radius of the ring: \(r=3\) = \(0.03\)
A thin-film interference pattern changes when the optical path difference changes by half a wavelength in the medium:
\(∆t=\frac{\lambda }{2\mu }\)
This means that the thickness decreases by:
\(\Delta t=\frac{\lambda }{2\mu }\)
Substituting values:
\(\Delta t=\frac{580\times 1{0}^{−9}}{2\times 1.45}\)
\(\Delta t=\frac{580\times 1{0}^{−9}}{2.9}\)
\(\Delta t=200\times 1{0}^{−9}\text{ m}=2\times 1{0}^{−7}\text{ m}\)
This thickness decreases every 12 seconds, so the rate of evaporation per second:
\(\frac{∆t}{dt}=\frac{2\times 1{0}^{−7}\text{ m}}{12}\) \(=1.67\times 1{0}^{−8}\text{ m/s}\)
Step 3: Finding the Volume Evaporation RateThe evaporating volume per second is:
\(\text{Rate of evaporation}=\text{Surface area}\times \frac{∆t}{dt}\)\(=\pi {r}^{2}\times \frac{∆t}{dt}\)
Substituting values:
\(=\pi (0.03{)}^{2}\times (1.67\times 1{0}^{−8})\)
\(=\pi (9\times 1{0}^{−4})\times (1.67\times 1{0}^{−8})\)
\(=15\pi \times 1{0}^{−12}{\text{ m}}^{3}\mathrm{/}\text{s}\)
Step 4: Choosing the Correct OptionFrom the given options, the correct answer : \(15\pi \times 1{0}^{−12}\)
A Young's double-slit experimental set up is kept in a medium of refractive index \(\left(\frac{4}{3}\right)\). Which maximum in this case will coincide with the \({6}^{\text{th }}\) maximum obtained if the medium is replaced by air?
\({8}^{\text{th }}\)
For the \({6}^{th}\)Sv6Kpe[] maximum in air to coincide with the \({n}^{th}\)Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[]
- Sv6Kpe[]
Sv6Kpe[]
\({y}_{6,air}=\frac{6{\lambda }_{air}D}{d}\) \({y}_{n,medium}=\frac{n{\lambda }_{medium}D}{d}=\frac{n(\frac{{\lambda }_{air}}{4/3})D}{d}\) Solving for \(n\)Sv6Kpe[]Sv6Kpe[]: Setting the two equations equal:Sv6Kpe[] \(\frac{6{\lambda }_{air}D}{d}=\frac{n(\frac{{\lambda }_{air}}{4/3})D}{d}\) \(6=\frac{n}{(4/3)}\) \(6=n\times \frac{3}{4}\) \(n=6\times \frac{4}{3}=8\)A glass slab of refractive index \({\mu }_{0}=1.44\) is coated with a thin film of refractive index \({\mu }_{f}=2\). The minimum thickness of the film, so that maximum transmission of green light of wavelength λ = 5000A˚ (incident normally) takes place, is:
(Shift II Memory Based)
1250A˚
To achieve maximum transmission of green light, the thin film must cause destructive interference for the light reflected from its top and bottom surfaces. This is a standard case of thin-film interference, and the condition for minimum thickness for maximum transmission (destructive reflection) is:
\(2\mathrm{n}\mathrm{t}=\mathrm{m}\lambda\)
where: n is the refractive index of the film, t is the thickness of the film, \(\lambda\) is the wavelength of light in the medium of the film, m is the order of interference (minimum for the thinnest film).
Here, m = 1
So, \(\mathrm{t}=\frac{\mathrm{m}\lambda }{2\mathrm{n}}=\frac{1\times 5000}{2\times 2}=1250\overset{^\circ }{\mathrm{A}}\)
In a Young's Double Slit Experiment (YDSE), for a wavelength \({\lambda }_{1}=600nm\), the 10th bright fringe is observed at a distance of 10 mm from the central maximum.For a new wavelength \({\lambda }_{2}=660nm\), what will be the distance of the 10th bright fringe from the central maximum?
(Shift - I Memory based)
Options are free to see. Unlock the correct answer and full explanation with Pass.
A ring of radius 3 cm has a soap film which is getting evaporated. Light of wavelength \(\lambda=580 \mathrm{~nm}\) gives minimum transmission every 12 s . Find the rate of evaporation. (Refractive index=1.45)(Shift - II Memory Based)
Options are free to see. Unlock the correct answer and full explanation with Pass.
The width of one of the two slits in Young's double slit experiment is d while that of the other slit is \(x\mathrm{d}\). If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is \(9:4\) then what is the value of \(x\) ?
(Assume that the field strength varies according to the slit width.)
[JEE Main 2025, 23 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
The width of one of the two slits in Young's double slit experiment is d while that of the other slit is \(x\mathrm{d}\). If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is \(9:4\) then what is the value of \(x\) ?
(Assume that the field strength varies according to the slit width.)
[JEE Main 2025, 23 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
In a Young's double-slit experiment, the fringe width is found to be \(\beta\). If the entire apparatus is immersed in a liquid of refractive index \(\mu\), the new fringe width will be :
Options are free to see. Unlock the correct answer and full explanation with Pass.
A plane wavefront is incident on a concave mirror of radius of curvature \(\mathrm{R}\). The radius of the refracted wavefront will be :
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two slits in Young's double slit experiment are \(1.5\mathrm{mm}\) apart and the screen is placed at a distance of \(1\mathrm{m}\) from the slits. If the wavelength of light used is \(600\times {10}^{-9}\mathrm{m}\) then the fringe separation is :
[Re-NEET 2024]
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the monochromatic source in Young's double slit experiment is replaced by white light, then
[NEET 2024]
Options are free to see. Unlock the correct answer and full explanation with Pass.
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is
Options are free to see. Unlock the correct answer and full explanation with Pass.
Light emerges out of a convex lens when a source of light kept at its focus. The shape of wavefront of the light is :
Options are free to see. Unlock the correct answer and full explanation with Pass.
A glass slab of refractive index \({\mu }_{0}=1.44\) is coated with a thin film of refractive index \({\mu }_{f}=2\). The minimum thickness of the film, so that maximum transmission of green light of wavelength λ = 5000A˚ (incident normally) takes place, is:
(Shift II Memory Based)
Options are free to see. Unlock the correct answer and full explanation with Pass.
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is
Options are free to see. Unlock the correct answer and full explanation with Pass.
A Young's double-slit experimental set up is kept in a medium of refractive index \(\left(\frac{4}{3}\right)\). Which maximum in this case will coincide with the \({6}^{\text{th }}\) maximum obtained if the medium is replaced by air?
Options are free to see. Unlock the correct answer and full explanation with Pass.
Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m, will be :
Options are free to see. Unlock the correct answer and full explanation with Pass.
A Young's double-slit experimental set up is kept in a medium of refractive index \(\left(\frac{4}{3}\right)\). Which maximum in this case will coincide with the \({6}^{\text{th }}\) maximum obtained if the medium is replaced by air?
Options are free to see. Unlock the correct answer and full explanation with Pass.
When unpolarized light is incident at an angle of \(60^{\circ}\) on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is:
Options are free to see. Unlock the correct answer and full explanation with Pass.
A Young's double-slit experimental set up is kept in a medium of refractive index \(\left(\frac{4}{3}\right)\). Which maximum in this case will coincide with the \({6}^{\text{th }}\) maximum obtained if the medium is replaced by air?
Options are free to see. Unlock the correct answer and full explanation with Pass.
Assertion (A) : In a Young's double-slit experiment, interference pattern is not observed when two coherent sources are infinitely close to each other.
Reason (R) : The fringe width is proportional to the separation between the two sources.
Options are free to see. Unlock the correct answer and full explanation with Pass.
In the wave picture of light, the intensity I of light is related to the amplitude \(A\) of the wave as :
Options are free to see. Unlock the correct answer and full explanation with Pass.
In a Young's double slit experiment, two slits are separated by \(2 mm\) and the screen is placed one meter away. When a light of wavelength \(500 nm\) is used, the fringe separation will be:
Options are free to see. Unlock the correct answer and full explanation with Pass.
According to Huygens principle, the amplitude of secondary wavelets is
Options are free to see. Unlock the correct answer and full explanation with Pass.
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, then the fringe width becomes
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the monochromatic source in Young's double slit experiment is replaced by white light, then
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two slits in Young's double slit experiment are \(1.5\mathrm{mm}\) apart and the screen is placed at a distance of \(1\mathrm{m}\) from the slits. If the wavelength of light used is \(600\times {10}^{-9}\mathrm{m}\) then the fringe separation is :
Options are free to see. Unlock the correct answer and full explanation with Pass.
For questions , two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :
Assertion (A) : In interference and diffraction of light, light energy reduces in one region producing a dark fringe. It increases in another region and produces a bright fringe.
Reason (R) : This happens because energy is not conserved in the phenomena of interference and diffraction.
Options are free to see. Unlock the correct answer and full explanation with Pass.
Assertion (A) : In Young's double slit experiment all fringes are of equal width.
Reason (R) : The fringe width depends upon wavelength of light \((\lambda)\) used, distance of screen from plane of slits (D) and slits separation (d).
Options are free to see. Unlock the correct answer and full explanation with Pass.
In a Young's double slits experiment, the ratio of amplitude of light coming from slits is \(2: 1\). The ratio of the maximum to minimum intensity in the interference pattern is :
[JEE Main 2023, 13 Apr (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
Assertion (A): The phase difference between any two points on a wavefront is zero.
Reason (R): All points on a wavefront are at the same distance from the source and thus oscillate in the same phase.
Options are free to see. Unlock the correct answer and full explanation with Pass.
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, then the fringe width becomes :
Options are free to see. Unlock the correct answer and full explanation with Pass.
The Brewsters angle \({i}_{b}\)for an interface should be :
Options are free to see. Unlock the correct answer and full explanation with Pass.
Practice more Board Physics PYQs
Browse every Physics chapter, or explore the full Board question bank.
All Physics chapters →