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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
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
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 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
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)
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
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
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 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
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.
The Young’s double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then [1990]- a)x (blue) = x (green)
- b)x (blue) > x (green)
- c)x (blue) < x (green)
- d)
Correct answer is option 'C'. Can you explain this answer?
The Young’s double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then [1990]
a)
x (blue) = x (green)
b)
x (blue) > x (green)
c)
x (blue) < x (green)
d)
|
Maitri Mukherjee answered |
Distance of nth maxima, 
Chapter doubts & questions for Wave Optics - NEET Past Year Papers 2025 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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