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Light of wavelength 500 nm is incident on a slit of width 0.1 mm. The width of the central bright line on the screen is 2m. What is the distance of the screen?a)1 mb)200 mc)0.75 md)0.5 mCorrect answer is option 'B'. Can you explain this answer?
|
Swara Sharma answered |
Beta(central Maxima)=2lambda D(distance of the screen)/d(distance between slits).
beta=2m.
wavelength= 500nm=5×10^-7m.
d=0.1mm=1×10^-4m.
2=2×5×10^-7 D/10^-4.
1=5×10^-3 D .
1/5×10^-3=D .
1000/5=D.
200m=D
beta=2m.
wavelength= 500nm=5×10^-7m.
d=0.1mm=1×10^-4m.
2=2×5×10^-7 D/10^-4.
1=5×10^-3 D .
1/5×10^-3=D .
1000/5=D.
200m=D
If the Young’s apparatus is immersed in water, the effect on fringe width will- a)remain same
- b)decrease
- c)increase
- d)Cant say because the experiment cannot be carried in any other medium except air
Correct answer is option 'B'. Can you explain this answer?
If the Young’s apparatus is immersed in water, the effect on fringe width will
a)
remain same
b)
decrease
c)
increase
d)
Cant say because the experiment cannot be carried in any other medium except air
|
Kalyan Joshi answered |
Effect of Immersing Young's Apparatus in Water on Fringe Width
When the Young's apparatus is immersed in water, the effect on the fringe width can be explained as follows:
1. Refractive Index of Water: Water has a higher refractive index than air. This means that light passing through water will be bent more than in air.
2. Path Difference: Path difference is the difference in the distance traveled by two waves from the source to the point where they interfere with each other. When the apparatus is immersed in water, the path difference between the two waves will change due to the change in refractive index.
3. Decrease in Fringe Width: As the path difference changes, the interference pattern will also change. The fringe width will decrease because the change in the path difference will cause the interference pattern to become more spread out.
4. Explanation: The fringe width is inversely proportional to the square root of the refractive index. Since the refractive index of water is higher than air, the fringe width will decrease when the apparatus is immersed in water.
Conclusion
In conclusion, when the Young's apparatus is immersed in water, the fringe width will decrease due to the change in refractive index, which leads to a change in the path difference and the interference pattern.
When the Young's apparatus is immersed in water, the effect on the fringe width can be explained as follows:
1. Refractive Index of Water: Water has a higher refractive index than air. This means that light passing through water will be bent more than in air.
2. Path Difference: Path difference is the difference in the distance traveled by two waves from the source to the point where they interfere with each other. When the apparatus is immersed in water, the path difference between the two waves will change due to the change in refractive index.
3. Decrease in Fringe Width: As the path difference changes, the interference pattern will also change. The fringe width will decrease because the change in the path difference will cause the interference pattern to become more spread out.
4. Explanation: The fringe width is inversely proportional to the square root of the refractive index. Since the refractive index of water is higher than air, the fringe width will decrease when the apparatus is immersed in water.
Conclusion
In conclusion, when the Young's apparatus is immersed in water, the fringe width will decrease due to the change in refractive index, which leads to a change in the path difference and the interference pattern.
A rotating calcite crystal is placed over an ink dot. On seeing through the crystal one finds:- a)Two stationary dots
- b)Two dots moving along straight lines
- c)One dot rotating about the other
- d)Both dots rotating about a common axis
Correct answer is option 'C'. Can you explain this answer?
A rotating calcite crystal is placed over an ink dot. On seeing through the crystal one finds:
a)
Two stationary dots
b)
Two dots moving along straight lines
c)
One dot rotating about the other
d)
Both dots rotating about a common axis
|
Anaya Patel answered |
In calcite crystal there is a phenomenon of double refraction. So, we see one dot as a stationary object whereas the other being refracted at a little displaced position seems to be rotating about the first dot.
An unpolarised beam of intensity Io is incident on a polarizer and analyser placed in contact. The angle between the transmission axes of the polarizer and the analyser is θ. What is the intensity of light emerging out of the analyser?- a)(I0 cos 2θ)/2
- b)I0 cos2θ
- c)I0 cosθ
- d)I0
Correct answer is option 'A'. Can you explain this answer?
An unpolarised beam of intensity Io is incident on a polarizer and analyser placed in contact. The angle between the transmission axes of the polarizer and the analyser is θ. What is the intensity of light emerging out of the analyser?
a)
(I0 cos 2θ)/2
b)
I0 cos2θ
c)
I0 cosθ
d)
I0
|
Nikita Singh answered |
Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E0 is the amplitude of the electric vector transmitted by the polarizer, then intensity I0 of the light incident on the analyser is
I ∞ E02
The electric field vector E0 can be resolved into two rectangular components i.e E0 cosθ and E0 sinθ. The analyzer will transmit only the component ( i.e E0 cosθ ) which is parallel to its transmission axis. However, the component E0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
I ∞ ( E0 x cosθ )2
I / I0 = ( E0 x cosθ )2 / E02 = cos2θ
I = I0 x cos2θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I0 x cos2θ/2
I ∞ E02
The electric field vector E0 can be resolved into two rectangular components i.e E0 cosθ and E0 sinθ. The analyzer will transmit only the component ( i.e E0 cosθ ) which is parallel to its transmission axis. However, the component E0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
I ∞ ( E0 x cosθ )2
I / I0 = ( E0 x cosθ )2 / E02 = cos2θ
I = I0 x cos2θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I0 x cos2θ/2
Two coherent sources produce a dark fringe when phase difference between the interfering waves is(n integer)- a)2π
- b)(2n – 1)π
- c)zero
- d)n
Correct answer is option 'B'. Can you explain this answer?
Two coherent sources produce a dark fringe when phase difference between the interfering waves is(n integer)
a)
2π
b)
(2n – 1)π
c)
zero
d)
n
|
Shruti Sarkar answered |
Dark fringes will be produced when there are destructive interference. The condition for that is the two waves should have a phase difference of an odd integral multiple of π.
Diffraction of light gives the information of- a)Transverse nature
- b)Longitudinal nature
- c)Both transverse and longitudinal
- d)Neither transverse nor longitudina
Correct answer is option 'D'. Can you explain this answer?
Diffraction of light gives the information of
a)
Transverse nature
b)
Longitudinal nature
c)
Both transverse and longitudinal
d)
Neither transverse nor longitudina
|
Vijay Bansal answered |
Transverse and longitudinal waves. Light and other types of electromagnetic radiation are transverse waves. Water waves and S waves (a type of seismic wave) are also transverse waves. In transverse waves, the vibrations are at right angles to the direction of travel.
Diffraction pattern cannot be observed with:a)one wide slitb)two narrow slitsc)one narrow slitd)large number of narrow slitsCorrect answer is option 'A'. Can you explain this answer?
|
Pooja Mehta answered |
If one illuminates two parallel slits, the light from the two slits again interferes. Here the interference is a more pronounced pattern with a series of alternating light and dark bands. ... He also proposed (as a thought experiment) that if detectors were placed before each slit, the interference pattern would disappear.
In a single slit experiment, suppose the slit width is equal to the wavelength of light used. Then- a)Slit image becomes very small
- b)The image of the slit becomes infinitely wide
- c)Diffraction effect cannot be observed
- d)Diffraction pattern becomes wider
Correct answer is option 'B'. Can you explain this answer?
In a single slit experiment, suppose the slit width is equal to the wavelength of light used. Then
a)
Slit image becomes very small
b)
The image of the slit becomes infinitely wide
c)
Diffraction effect cannot be observed
d)
Diffraction pattern becomes wider
|
Om Desai answered |
This is because for maximum diffraction the slit width always equals to the wavelength of light.
The phenomena diffraction can take place in sound waves.- a)Yes
- b)No
- c)Only Interference
- d)Under certain conditions only
Correct answer is option 'A'. Can you explain this answer?
The phenomena diffraction can take place in sound waves.
a)
Yes
b)
No
c)
Only Interference
d)
Under certain conditions only
|
Rohan Singh answered |
YES, Sound waves, on the other hand, are longitudinal, meaning that they oscillate parallel to the direction of their motion. Since there is no component of a sound wave's oscillation that is perpendicular to its motion, sound waves cannot be polarized
Which of the following undergoes largest diffraction?- a)Ultraviolet light
- b)Infra red light
- c)Radio waves
- d)Y – rays
Correct answer is option 'C'. Can you explain this answer?
Which of the following undergoes largest diffraction?
a)
Ultraviolet light
b)
Infra red light
c)
Radio waves
d)
Y – rays
|
|
Anjana Sharma answered |
Maximum diffraction occurs when size of obstacle is almost equal to wavelength of light wave. Hence maximum diffraction occurs for larger wavelength . As wavelength of radio wave is higher than others, maximum diffraction will occur for it.
Young’s double slit experiment is carried out using two bulbs instead of using two slits and one source. Then- a)Fringes will be elliptical
- b)No interference fringes will be observed
- c)Fringes will be dark
- d)The interference fringes will be colored
Correct answer is option 'B'. Can you explain this answer?
Young’s double slit experiment is carried out using two bulbs instead of using two slits and one source. Then
a)
Fringes will be elliptical
b)
No interference fringes will be observed
c)
Fringes will be dark
d)
The interference fringes will be colored
|
|
Rajesh Gupta answered |
We observe interference pattern when the size of the slits are comparable to the wavelength of light. But here, the size of the bulb is bigger than wavelength of light.
A ray of light strikes a glass plate at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, then refractive index of glass is- a)3/2
- b)√3
- c)√3/2
- d)1/2
Correct answer is option 'B'. Can you explain this answer?
A ray of light strikes a glass plate at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, then refractive index of glass is
a)
3/2
b)
√3
c)
√3/2
d)
1/2
|
Hansa Sharma answered |
Given from figure angle of incidence i=600
angle of refraction r=300
We know μ= sin i/sin r
μ= sin60/sin30 =((√3)/2) ×2=√3
In Young’s double slit experiment, the distance between the two slits is halved and the distance of the screen from the slit is doubled. The fringe width will be- a)Will become four times
- b)Doubled
- c)Remain same
- d)Halved
Correct answer is option 'A'. Can you explain this answer?
In Young’s double slit experiment, the distance between the two slits is halved and the distance of the screen from the slit is doubled. The fringe width will be
a)
Will become four times
b)
Doubled
c)
Remain same
d)
Halved
|
|
Anaya Patel answered |
Fringe width β=λD/d
On d′=d/2, D′=2D
New fringe width β′= λD′/d′=4β
On d′=d/2, D′=2D
New fringe width β′= λD′/d′=4β
When viewed in white light, a soap bubble shows colours because of- a)Interference
- b)Scattering
- c)Dispersion
- d)Diffraction
Correct answer is option 'A'. Can you explain this answer?
When viewed in white light, a soap bubble shows colours because of
a)
Interference
b)
Scattering
c)
Dispersion
d)
Diffraction
|
Nayanika Datta answered |
Soap Bubble and Colors
Introduction:
When white light passes through a soap bubble, it undergoes a phenomenon known as dispersion, resulting in the appearance of various colors. This colorful phenomenon is due to interference of light waves and can be explained by the principles of interference and thin film interference.
Interference:
Interference is the interaction of two or more light waves, resulting in either constructive or destructive interference. Constructive interference occurs when the peaks and troughs of the waves align, leading to reinforcement and the formation of bright regions. Destructive interference occurs when the peaks and troughs of the waves cancel each other out, resulting in dark regions.
Thin Film Interference:
A soap bubble can be considered as a thin film of water sandwiched between two layers of air. When light waves hit the surface of the soap bubble, a portion of the light is reflected, and another portion enters the bubble. This incident light can also be seen as a combination of different wavelengths, or colors, of light.
Multiple Reflections and Interference:
As the light waves enter the soap bubble, they undergo multiple reflections from the inner and outer surfaces of the film. These reflections create a path difference between the different waves. If the path difference is an integer multiple of the wavelength of light, constructive interference occurs, resulting in bright colors. If the path difference is a half-integer multiple of the wavelength, destructive interference occurs, leading to dark regions.
Dispersion:
In addition to interference, dispersion also plays a role in the formation of colors in soap bubbles. Dispersion is the phenomenon where different colors of light have different wavelengths and therefore different speeds when passing through a medium. As the light waves enter the soap film, they are separated into their constituent colors due to their different wavelengths. This separation of colors is what gives rise to the vibrant and distinct colors observed in soap bubbles.
Conclusion:
In conclusion, soap bubbles show colors when viewed in white light due to the phenomenon of interference and dispersion. The interaction of light waves in the thin film of the bubble, combined with the separation of colors based on their wavelengths, leads to the formation of various colors. This colorful display is a result of the complex interplay between interference and dispersion.
Introduction:
When white light passes through a soap bubble, it undergoes a phenomenon known as dispersion, resulting in the appearance of various colors. This colorful phenomenon is due to interference of light waves and can be explained by the principles of interference and thin film interference.
Interference:
Interference is the interaction of two or more light waves, resulting in either constructive or destructive interference. Constructive interference occurs when the peaks and troughs of the waves align, leading to reinforcement and the formation of bright regions. Destructive interference occurs when the peaks and troughs of the waves cancel each other out, resulting in dark regions.
Thin Film Interference:
A soap bubble can be considered as a thin film of water sandwiched between two layers of air. When light waves hit the surface of the soap bubble, a portion of the light is reflected, and another portion enters the bubble. This incident light can also be seen as a combination of different wavelengths, or colors, of light.
Multiple Reflections and Interference:
As the light waves enter the soap bubble, they undergo multiple reflections from the inner and outer surfaces of the film. These reflections create a path difference between the different waves. If the path difference is an integer multiple of the wavelength of light, constructive interference occurs, resulting in bright colors. If the path difference is a half-integer multiple of the wavelength, destructive interference occurs, leading to dark regions.
Dispersion:
In addition to interference, dispersion also plays a role in the formation of colors in soap bubbles. Dispersion is the phenomenon where different colors of light have different wavelengths and therefore different speeds when passing through a medium. As the light waves enter the soap film, they are separated into their constituent colors due to their different wavelengths. This separation of colors is what gives rise to the vibrant and distinct colors observed in soap bubbles.
Conclusion:
In conclusion, soap bubbles show colors when viewed in white light due to the phenomenon of interference and dispersion. The interaction of light waves in the thin film of the bubble, combined with the separation of colors based on their wavelengths, leads to the formation of various colors. This colorful display is a result of the complex interplay between interference and dispersion.
In the diffraction of a single slit experiment, if the slit width is half then the width of the central maxima will be- a)Halved
- b)Becomes four times.
- c)Remains the same
- d)Doubled
Correct answer is option 'D'. Can you explain this answer?
In the diffraction of a single slit experiment, if the slit width is half then the width of the central maxima will be
a)
Halved
b)
Becomes four times.
c)
Remains the same
d)
Doubled
|
Amit Kumar answered |
Fringe width is inversly prop to dist betweenslit so it is double
The Brewster’s angle for a transparent medium is 600.The angle of incidence is- a)450
- b)900
- c)600
- d)300
Correct answer is option 'C'. Can you explain this answer?
The Brewster’s angle for a transparent medium is 600.The angle of incidence is
a)
450
b)
900
c)
600
d)
300
|
|
Hansa Sharma answered |
Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.
So the angle of incidence would be 60°
So the angle of incidence would be 60°
The intensity of light transmitted by the analyzer is maximum when- a)The transmission axes of the analyzer and the polarizer are perpendicular
- b)The reflected ray is fully polarized
- c)The transmission axes of the analyzer and the polarizer are parallel
- d)The reflected ray is partially polarized
Correct answer is option 'C'. Can you explain this answer?
The intensity of light transmitted by the analyzer is maximum when
a)
The transmission axes of the analyzer and the polarizer are perpendicular
b)
The reflected ray is fully polarized
c)
The transmission axes of the analyzer and the polarizer are parallel
d)
The reflected ray is partially polarized
|
|
Alfiya I answered |
From malu's law,
If I0 is intensity of plane polarised light incident on an analyser and (thete) is the angle between axes of polariser and analyser, then intensity of polarised light transmitted from analyser ,I = I0 cos(^2) thete,
ie, when (thete)=0, cos(^2) 0=1, then I= I0
ie, the transmission axes of polariser and analyser are paralell.
What is the angular spread of the first minimum and the central maximum when diffraction is observed with 540 nm light through a slit of width 1mm?- a)540 x 10-6 rad
- b)1080 x 10-6 rad
- c)270 x 10-6 rad
- d)810 x 10-6 rad
Correct answer is option 'D'. Can you explain this answer?
What is the angular spread of the first minimum and the central maximum when diffraction is observed with 540 nm light through a slit of width 1mm?
a)
540 x 10-6 rad
b)
1080 x 10-6 rad
c)
270 x 10-6 rad
d)
810 x 10-6 rad
|
|
Rahul Kapoor answered |
The condition for the first minimum ,
So the ? is the angular separation between first minimum and the central maximum.
So the correct answer is (A) . 540 x 10-6 rad
The change in diffraction pattern of a single slit, when the monochromatic source of light is replaced by a source of white light will be- a)Diffracted image is not clear
- b)Diffracted image gets dispersed into constituent colours of white light
- c)The image of the slit becomes infinitely wide
- d)Clear colourful fringe pattern
Correct answer is option 'B'. Can you explain this answer?
The change in diffraction pattern of a single slit, when the monochromatic source of light is replaced by a source of white light will be
a)
Diffracted image is not clear
b)
Diffracted image gets dispersed into constituent colours of white light
c)
The image of the slit becomes infinitely wide
d)
Clear colourful fringe pattern
|
Dr Manju Sen answered |
- When a source of white light is used instead of a monochromatic source, the diffracted image of the slit gets dispersed into constituent colours of white light.
- The central maximum will be white and on either side of the central maximum, there will be coloured fringes.
In the Young’s double slit experiment, the distance of p th dark fringe from the central maximum is:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In the Young’s double slit experiment, the distance of p th dark fringe from the central maximum is:
a)
b)
c)
d)
|
|
Neha Sharma answered |
Position of the pth dark fringe is at a distance of
y=(2p−1) (λD/2d) from the centre
where
λ: Wavelength
D: Distance between slits
d: slit width
y=(2p−1) (λD/2d) from the centre
where
λ: Wavelength
D: Distance between slits
d: slit width
In an interference experiment monochromatic light is replaced by white light, we will see:- a)uniform darkness on the screen
- b)a few coloured bands and then uniform illumination
- c)equally spaced white and dark bands
- d)uniform illumination on the screen
Correct answer is option 'B'. Can you explain this answer?
In an interference experiment monochromatic light is replaced by white light, we will see:
a)
uniform darkness on the screen
b)
a few coloured bands and then uniform illumination
c)
equally spaced white and dark bands
d)
uniform illumination on the screen
|
|
Arun Khanna answered |
Therefore if monochromatic light in Young's interference experiment is replaced by white light, then the waves of each wavelength form their separate interference patterns. The resultant effect of all these patterns is obtained on the screen. i.e., the waves of all colours reach at mid point M in same phase we will see.a few coloured bands and then uniform illumination
If the light is completely polarized by reflection, then angle between the reflected and refracted light is- a)π
- b)2π
- c)π/2
- d)π/4
Correct answer is option 'C'. Can you explain this answer?
If the light is completely polarized by reflection, then angle between the reflected and refracted light is
a)
π
b)
2π
c)
π/2
d)
π/4
|
Sant Singh answered |
This is in accordance with Brewsters law. When light is incident. When light is incident at Brewsters angle then the refracted and reflected ray are perpendicular to reach other.
In Young’s double slit experiment, the slits are 2 mmapart and are illuminated by photons of twowavelengths λ1 = 12000Å and λ2 = 10000Å. Atwhat minimum distance from the common centralbright fringe on the screen 2 m from the slit will abright fringe from one interference patterncoincide with a bright fringe from the other ? [NEET 2013]- a)6 mm
- b)4 mm
- c)3 mm
- d)8 mm
Correct answer is option 'A'. Can you explain this answer?
In Young’s double slit experiment, the slits are 2 mmapart and are illuminated by photons of twowavelengths λ1 = 12000Å and λ2 = 10000Å. Atwhat minimum distance from the common centralbright fringe on the screen 2 m from the slit will abright fringe from one interference patterncoincide with a bright fringe from the other ? [NEET 2013]
a)
6 mm
b)
4 mm
c)
3 mm
d)
8 mm
|
Anu Bajaj answered |
∴ n1 λ1 = n2λ2
⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
⇒n = 5
⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
⇒n = 5
(∵ λ1 = 1200 x 10-10m; λ2 = 10000 x 10-10m)
(∵ d = 2 mm and D = 2m)
= 5 × 12 × 10–4 m
= 60 × 10–4 m
= 6 × 10–3m = 6 mm
= 5 × 12 × 10–4 m
= 60 × 10–4 m
= 6 × 10–3m = 6 mm
A ray of light is incident on the surface of glass plate at an angle of incidence equal to Brewster’s angle Ø. If μ represents the refractive index of glass with respect to air, then the angle between the reflected and refracted rays is:- a)90°- sin-1 (sinØμ)
- b)90 + Ø
- c)sin-1 (μ cosØ)
- d)90°
Correct answer is option 'D'. Can you explain this answer?
A ray of light is incident on the surface of glass plate at an angle of incidence equal to Brewster’s angle Ø. If μ represents the refractive index of glass with respect to air, then the angle between the reflected and refracted rays is:
a)
90°- sin-1 (sinØμ)
b)
90 + Ø
c)
sin-1 (μ cosØ)
d)
90°
|
Jayant Mishra answered |
In Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refrative index 4/3, without disturbing the geometrical arrangement, the new fringe width will be [1990]- a)0.30 mm
- b)0.40 mm
- c)0.53 mm
- d)450 microns
Correct answer is option 'A'. Can you explain this answer?
In Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refrative index 4/3, without disturbing the geometrical arrangement, the new fringe width will be [1990]
a)
0.30 mm
b)
0.40 mm
c)
0.53 mm
d)
450 microns
|
Tejas Chavan answered |
A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]- a)0°
- b)15°
- c)30°
- d)50°
Correct answer is option 'C'. Can you explain this answer?
A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]
a)
0°
b)
15°
c)
30°
d)
50°
|
|
Anu Bajaj answered |
For first minimum, a sin θ = nλ = 1λ
A paper, with two marks having separation d, isheld normal to the line of sight of an observer ata distance of 50m. The diameter of the eye-lensof the observer is 2 mm. Which of the followingis the least value of d, so that the marks can beseen as separate ? The mean wavelength ofvisible light may be taken as 5000 Å. [2002]
- a)1.25 m
- b)13.6 cm
- c)12.5 cm
- d)2.5 mm
Correct answer is option 'C'. Can you explain this answer?
A paper, with two marks having separation d, isheld normal to the line of sight of an observer ata distance of 50m. The diameter of the eye-lensof the observer is 2 mm. Which of the followingis the least value of d, so that the marks can beseen as separate ? The mean wavelength ofvisible light may be taken as 5000 Å. [2002]
a)
1.25 m
b)
13.6 cm
c)
12.5 cm
d)
2.5 mm
|
Gowri Nair answered |
Angular limit of resolution of eye, 
where, d is diameter of eye lens.
Also, if y is the minimum separation between two objects at distance D from eye then
Also, if y is the minimum separation between two objects at distance D from eye then
θ = y/D
Here, wavelength λ = 5000Å = 5x10-7m
D = 50 m
Diameter of eye lens = 2 mm = 2x10-3 m
From eq. (1), minimum separation is
D = 50 m
Diameter of eye lens = 2 mm = 2x10-3 m
From eq. (1), minimum separation is
y = 5 * 10-7 * 50 / 2 * 10-3
y = 0.0125 m
y = 1.25 cm
In Young’s experiment, two coherent sources areplaced 0.90 mm apart and fringe are observedone metre away. If it produces second dark fringeat a distance of 1 mm from central fringe, thewavelength of monochromatic light used wouldbe [1991, 1992]- a)60 × 10–4 cm
- b)10 × 10–4 cm
- c)10 × 10–5 cm
- d)6 × 10–5 cm
Correct answer is option 'D'. Can you explain this answer?
In Young’s experiment, two coherent sources areplaced 0.90 mm apart and fringe are observedone metre away. If it produces second dark fringeat a distance of 1 mm from central fringe, thewavelength of monochromatic light used wouldbe [1991, 1992]
a)
60 × 10–4 cm
b)
10 × 10–4 cm
c)
10 × 10–5 cm
d)
6 × 10–5 cm
|
Devansh Mehra answered |
For dark fringe
λ = 0.6x10-6 m = 6 × 10–5 cm
Which of the following phenomenon is notcommon to sound and light waves ? [1988]- a)Interference
- b)Diffraction
- c)Coherence
- d)Polarisation
Correct answer is option 'D'. Can you explain this answer?
Which of the following phenomenon is notcommon to sound and light waves ? [1988]
a)
Interference
b)
Diffraction
c)
Coherence
d)
Polarisation
|
Shivani Rane answered |
Sound waves can not be polarised as they are longitudinal. Light waves can be polarised as they are transverse.
In Young’s double slit experiment, one slit is covered. The observed effect will be- a)intensities of the fringes will be reduced
- b)interference will not take place
- c)colored fringes will be produced
- d)dark fringes will be replaced by bright ones
Correct answer is option 'B'. Can you explain this answer?
In Young’s double slit experiment, one slit is covered. The observed effect will be
a)
intensities of the fringes will be reduced
b)
interference will not take place
c)
colored fringes will be produced
d)
dark fringes will be replaced by bright ones
|
|
Divyansh Kulkarni answered |
Adult fiction, the term "In Young" is not a commonly used phrase or concept. It is possible that you may be referring to a specific book or series that uses this term in a unique way, but without further information, it is difficult to provide a more specific answer. Can you please provide more context or information about what you are referring to?
In Young’s double slit experiment carried out withlight of wavelength (λ) = 5000Å, the distancebetween the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zero th maximum) will be at x equal to [1992]- a)1.67 cm
- b)1.5 cm
- c)0.5 cm
- d)5.0 cm
Correct answer is option 'B'. Can you explain this answer?
In Young’s double slit experiment carried out withlight of wavelength (λ) = 5000Å, the distancebetween the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zero th maximum) will be at x equal to [1992]
a)
1.67 cm
b)
1.5 cm
c)
0.5 cm
d)
5.0 cm
|
Arya Khanna answered |
A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]- a)πλ
- b)2π
- c)3π
- d)4π
Correct answer is option 'D'. Can you explain this answer?
A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]
a)
πλ
b)
2π
c)
3π
d)
4π
|
Subhankar Datta answered |
Conditions for diffraction minima are
Path diff. Δx = nλ and Phase diff. δφ = 2nπ
Path diff. = nλ = 2λ
Phase diff. = 2nπ = 4π (∵n = 2)
Path diff. Δx = nλ and Phase diff. δφ = 2nπ
Path diff. = nλ = 2λ
Phase diff. = 2nπ = 4π (∵n = 2)
The device or arrangement to detect and check plane polarized light is called- a)Optically active
- b)Polaroid
- c)Polarizer
- d)Analyser
Correct answer is option 'D'. Can you explain this answer?
The device or arrangement to detect and check plane polarized light is called
a)
Optically active
b)
Polaroid
c)
Polarizer
d)
Analyser
|
|
Rajat Patel answered |
An analyser or analyzer is a person or device that analyses given data. It examines in detail the structure of the given data and tries to find patterns and relationships between parts of the data. An analyser can be a piece of hardware or a computer program running on a computer.
An unpolarised beam of intensity 2a2 passes through a thin polaroid. Assuming zero absorption in the polaroid, the intensity of emergent plane polarised light will be- a)a2
- b)2a2
- c)√2 a2
- d)a2 / √2
Correct answer is option 'A'. Can you explain this answer?
An unpolarised beam of intensity 2a2 passes through a thin polaroid. Assuming zero absorption in the polaroid, the intensity of emergent plane polarised light will be
a)
a2
b)
2a2
c)
√2 a2
d)
a2 / √2
|
Gunjan Lakhani answered |
Initial unpolarized intensity is 2a2
the intensity of light transmitted by the first polarizered will be Iunpolarized/2=a2
option A is correct
the intensity of light transmitted by the first polarizered will be Iunpolarized/2=a2
option A is correct
The diffraction of light is not easily noticed if- a)The wave length of visible range is small
- b)Light waves are transverse and longitudinal in nature
- c)Size of ordinary objects are small
- d)Apparatus are needed for noticing
Correct answer is option 'A'. Can you explain this answer?
The diffraction of light is not easily noticed if
a)
The wave length of visible range is small
b)
Light waves are transverse and longitudinal in nature
c)
Size of ordinary objects are small
d)
Apparatus are needed for noticing
|
Prakhar Maheshwari answered |
Actually, diffraction occurs all the time. But when the aperture/wavelength ratio is large, the spacing between fringes becomes too small, and the diffractive effects are unnoticeable.
Which one of the following phenomena is notexplained by Huygen’s construction ofwavefront? [1988]- a)Refraction
- b)Reflection
- c)Diffraction
- d)Origin of spectra
Correct answer is option 'D'. Can you explain this answer?
Which one of the following phenomena is notexplained by Huygen’s construction ofwavefront? [1988]
a)
Refraction
b)
Reflection
c)
Diffraction
d)
Origin of spectra
|
Krish Chakraborty answered |
Huygen's construction of wavefront does not apply to origin of spectra which is explained by quantum theory
Which of the following in conserved when light waves interfere?- a)Mass
- b)Amplitude
- c)Energy
- d)Intensity
Correct answer is option 'C'. Can you explain this answer?
Which of the following in conserved when light waves interfere?
a)
Mass
b)
Amplitude
c)
Energy
d)
Intensity
|
|
Rohan Singh answered |
A wave can transmit energy from one place to another.
Interference was observed in interferencechamber where air was present, now the chamberis evacuated, and if the same light is used, acareful observer will see [1993]- a)no interference
- b)interference with brighter bands
- c)interference with dark bands
- d)interference fringe with larger width
Correct answer is option 'D'. Can you explain this answer?
Interference was observed in interferencechamber where air was present, now the chamberis evacuated, and if the same light is used, acareful observer will see [1993]
a)
no interference
b)
interference with brighter bands
c)
interference with dark bands
d)
interference fringe with larger width
|
Pankaj Banerjee answered |
In vaccum, λ increases very slightly compared to that in air. As β∝ λ , therefore, width of interference fringe increases slightly.
The periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:- a)I1 + I2
- b)
- c)
- d)2(I1+ I2)
Correct answer is option 'D'. Can you explain this answer?
The periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:
a)
I1 + I2
b)
c)
d)
2(I1+ I2)
|
|
Tejas Chavan answered |
The resultant intensity of two periodic
waves at a point is given by
waves at a point is given by
Resultant intensity is maximum if
Resultant intensity is minimum if
Therefore, the sum of the maximum and
minimum intensities is Imax + Imin
minimum intensities is Imax + Imin
Interference is possible in [1989]- a)light waves only
- b)sound waves only
- c)both light and sound waves
- d)neither light nor sound waves
Correct answer is option 'C'. Can you explain this answer?
Interference is possible in [1989]
a)
light waves only
b)
sound waves only
c)
both light and sound waves
d)
neither light nor sound waves
|
Swara Desai answered |
Interference is a wave phenomenon shown by both the light waves and sound waves.
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter
2/d in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be
respectively:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter
2/d in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be
respectively:
2/d in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be
respectively:
a)
b)
c)
d)
|
|
Pankaj Banerjee answered |
By covering aperture, focal length does not change. But intensity is reduced by 1/4 times, as aperture diameter d/2 is covered.
∴ New focal length = f and intensity = 3I/4
A parallel beam of fast moving electrons isincident normally on a narrow slit. A fluorescentscreen is placed at a large distance from the slit.If the speed of the electrons is increased, whichof the following statements is correct ? [NEET 2013]- a)The angular width of the central maximumof the diffraction pattern will increase.
- b)The angular width of the central maximumwill decrease.
- c)The angular width of the central maximumwill be unaffected.
- d)Diffraction pattern is not observed on thescreen in case of electrons.
Correct answer is option 'B'. Can you explain this answer?
A parallel beam of fast moving electrons isincident normally on a narrow slit. A fluorescentscreen is placed at a large distance from the slit.If the speed of the electrons is increased, whichof the following statements is correct ? [NEET 2013]
a)
The angular width of the central maximumof the diffraction pattern will increase.
b)
The angular width of the central maximumwill decrease.
c)
The angular width of the central maximumwill be unaffected.
d)
Diffraction pattern is not observed on thescreen in case of electrons.
|
Vijay Gupta answered |
If the speed will increase then the angular width of central maximum will decrease as speed is inversely proportional to wavelength, so as wavelength decreases the angular width also decreases.
angular width =wavelength/distance between the slits.
angular width =wavelength/distance between the slits.
Which of the following shows particle nature oflight? [2001]- a)refraction
- b)interference
- c)polarization
- d)photoelectric effect
Correct answer is option 'D'. Can you explain this answer?
Which of the following shows particle nature oflight? [2001]
a)
refraction
b)
interference
c)
polarization
d)
photoelectric effect
|
|
Shivani Rane answered |
Photoelectric effect shows particle nature of light as light consists of packets of energy called photons or quanta
In Young’s double slit experiment the distancebetween the slits and the screen is doubled. Theseparation between the slits is reduced to half.As a result the fringe width [NEET Kar. 2013]- a)is doubled
- b)is halved
- c)becomes four times
- d)remains unchanged
Correct answer is option 'C'. Can you explain this answer?
In Young’s double slit experiment the distancebetween the slits and the screen is doubled. Theseparation between the slits is reduced to half.As a result the fringe width [NEET Kar. 2013]
a)
is doubled
b)
is halved
c)
becomes four times
d)
remains unchanged
|
Prasenjit Pillai answered |
Fringe width 
From question D' = 2D and d' = d/2
If yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by a monochromatic blue light of the same intensity [1992]- a)fringe width will decrease
- b)finge width will increase
- c)fringe width will remain unchanged
- d)fringes will become less intense
Correct answer is option 'A'. Can you explain this answer?
If yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by a monochromatic blue light of the same intensity [1992]
a)
fringe width will decrease
b)
finge width will increase
c)
fringe width will remain unchanged
d)
fringes will become less intense
|
|
Subhankar Datta answered |
As 
∴ Fringe width β will decrease
In a Fresnel biprism experiment, the two positionsof lens give separation between the slits as 16cm and 9 cm respectively. What is the actualdistance of separation? [1996]- a)12.5 cm
- b)12 cm
- c)13 cm
- d)14 cm
Correct answer is option 'B'. Can you explain this answer?
In a Fresnel biprism experiment, the two positionsof lens give separation between the slits as 16cm and 9 cm respectively. What is the actualdistance of separation? [1996]
a)
12.5 cm
b)
12 cm
c)
13 cm
d)
14 cm
|
|
Prasenjit Pillai answered |
Separation between slits are (r1=) 16 cm and (r2=) 9 cm.
Actual distance of separation
Actual distance of separation
Colours appear on a thin soap film and on soapbubbles due to the phenomenon of [1999]- a)refraction
- b)dispersion
- c)interference
- d)diffraction
Correct answer is option 'C'. Can you explain this answer?
Colours appear on a thin soap film and on soapbubbles due to the phenomenon of [1999]
a)
refraction
b)
dispersion
c)
interference
d)
diffraction
|
|
Devansh Mehra answered |
We know that the colours for which the
condition of constructive interference is
satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.
condition of constructive interference is
satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.
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