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LIGHT WAVES
EXERCISES Q-1 TO Q-10
1. Find the range of frequency of light that is visible to an average human being (400 nm < 𝜆 < 700 nm).
ANSWER: The velocity of light, c = 3x10⁸ m/s.
Since c = 𝜈𝜆
Hence 𝜈 = c/𝜆
for 𝜆 = 400 nm = 400x10⁻⁹ m
The frequency = 3x10⁸/400x10⁻⁹ Hz =(3/4)x10¹⁵ Hz
= 0.75 x10¹⁵ Hz = 7.5 x 10¹⁴ Hz
For 𝜆 = 700 nm = 700x10⁻⁹ m
The frequency = 3x10⁸/700x10⁻⁹ Hz =(3/7)x10¹⁵ Hz
=0.43x10¹⁵ Hz = 4.3x10¹⁴ Hz.
Hence the range of frequency of light that is visible to an average human being is
4.3x10¹⁴ Hz < 𝜈 <7.5x10¹⁴ Hz.
2. The wavelength of sodium light in air is 589 nm.
(a) Find its frequency in air.
(b) Find its wavelength in water (refractive index = 1.33).
(c) Find its frequency in water.
(d) Find its speed in water.
ANSWER: The wavelength of sodium light in air 𝜆=589 nm =589x10⁻⁹ m. The speed of light in air, c≈3x10⁸ m/s
(a) Hence its frequency in the air, 𝜈 = c/𝜆
=3x10⁸/589x10⁻⁹ Hz
=0.00509x10¹⁷ Hz
=5.09x10¹⁴ Hz
(b) Let the wavelength of sodium light in water =𝜆', and the speed of light in water c'. The frequency does not change, hence
c' =𝜈𝜆', and c = 𝜈𝜆
Thus, c'/c = 𝜆'/𝜆
{but c' = c/µ, where µ is the refractive index of water}
→c/µc = 𝜆'/𝜆
→1/µ = 𝜆'/𝜆
→𝜆' = 𝜆/µ =589/1.33 nm =443 nm
(c) When the light enters another medium, its wavelength changes not the frequency. Hence the frequency in water =𝜈 =5.09x10¹⁴ Hz
(d) The speed of sodium light in water c' = c/µ
=3x10⁸/1.33 m/s
=2.25 x10⁸ m/s
3. The index of refraction of fused quartz is 1.472 for the light of wavelength 400 nm and is 1.452 for the light of wavelength 760 nm. Find the speed of light of these wavelengths in fused quartz.
ANSWER: The speed of light in vacuum, c = 3x10⁸ m/s, The speed of light in the medium c' = c/µ, where µ is the index of refraction. For the light of wavelength 400 nm, given that µ = 1.472. Hence the speed of light for this wavelength
c' = 3x10⁸/1.472 m/s =2.04x10⁸ m/s
For the light of wavelength 750 nm, given that µ = 1.452. Hence the speed of light for this wavelength
c' = 3x10⁸/1.452 m/s =2.07x10⁸ m/s
4. The speed of the yellow light in a certain liquid is 2.4 x 10⁸ m/s. Find the refractive index of the liquid.
ANSWER: Since the speed of light in a medium, c' =c/µ. Where c = speed of light in vacuum and µ =refractive index of the medium. Here given that c' =2.4x10⁸ m/s and we know that c= 3x10⁸ m/s.
Hence, µ = c/c' =3x10⁸/2.4x10⁸ =1.25
5. Two narrow slits emitting light in phase are separated by a distance of 1.0 cm. The wavelength of the light is 5.0 x 10⁻⁷ m. The interference pattern is observed on a screen placed at a distance of 1.0 m.
(a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima?
(b) Find the separation between the sources which will give a separation of 1.0 mm between the consecutive maxima.
ANSWER: The separation of slits, d = 1.0 cm =0.01 m.
The wavelength of light, 𝜆 = 5.0x10⁻⁷ m.
The distance of the screen from the slits, D = 1.0 m.
(a) The separation between consecutive maxima = the fringe width, w = D𝜆/d =1.0*5.0x10⁻⁷/0.01 m
=5x10⁻⁵ *1000 mm =5x10⁻² mm =0.05 mm
The width is in the range of one-hundredth of a millimeter, hence they can hardly be distinguished.
Figure for Q-5
(b) Here given w = 1.0 mm = 0.001 m and d = ?
d = D𝜆/w =1.0*5.0x10⁻⁷/0.001 m
=5x10⁻⁴ *1000 mm =5/10 mm =0.50 mm
6. The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5 m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
ANSWER: Given fringe width, w = 1.0 mm =0.001 m, Distance of the screen, D = 2.5 m, the separation of slits, d = 1.0 mm =0.001 m. The wavelength 𝜆 =?
Since w =D𝜆/d
→𝜆 = wd/D =0.001*0.001/2.5 m =0.4x10⁻⁶ m
=4x10⁻⁷*10⁹ nm =400 nm
7. In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 x 10⁻⁷ m and the distance of the screen from the slits is 1.0 m.
(a) Find the distance of the center of the first minimum from the center of the central maximum.
(b) How many bright fringes are formed in one-centimeter width on the screen?
ANSWER: Given slit separation, d = 1.0 mm =0.001 m, The wavelength of the light, 𝜆 = 5.0x10⁻⁷ m and the distance of the screen, D = 1.0 m.
(a) Hence the fringe width, w =D𝜆/d
→w = 1.0*5.0x10⁻⁷/0.001 m =5x10⁻⁴*1000 mm
→w = 5/10 mm =0.50 mm.
This fringe width is the distance between the center of consecutive maxima. The distance of the center of the first minimum from the center of the central maximum = D𝜆/2d = 0.50/2 mm =0.25 mm
(b) The number of bright fringes in one-centimeter width = 10 mm/0.50 mm = 20
8. In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of the yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?
ANSWER: Given that, the separation of slits, d = 0.800 mm =8x10⁻⁴ m.
The wavelength of the light, 𝜆 = 589 nm =589x10⁻⁹ m.
The screen distance, D = 2.00 m. Fringe width, w =?W =D𝜆/d =2.00*589x10⁻⁹/8x10⁻⁴ m
=147.25*10⁻⁵*1000 mm
≈1.47 mm
9. Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is 2.0 x 10⁻³ m.
ANSWER: Given the slit separation, d = 2.0x10⁻³ m.
The wavelength of the light, 𝜆 =500 nm =500x10⁻⁹ m
Let, w = separation between the consecutive bright fringes.
D = The distance of the screen from the slits and
θ = angular separation between the consecutive bright fringes. Then,
θ = w/D =D𝜆/dD =𝜆/d =500x10⁻⁹/2.0x10⁻³ rad
→θ = 25x10⁻⁵*180°/π =0.014°
10. A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
ANSWER: Given that, the separation between the slits, d = 0.25 mm = 2.5x10⁻⁴ m
The distance of the screen from the slits, D = 150 cm =1.50 m
The first maximum occurs from the central maximum at a distance y =D𝜆/d, where 𝜆 is the wavelength of the light.
For the 𝜆 = 480 nm =480x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y = (D/d)*480x10⁻⁹ m
For the 𝜆' = 600 nm =600x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y' = (D/d)*600x10⁻⁹ m
Hence the linear separation between the first maximum (from the central maximum) corresponding to the two wavelengths =y'-y
=(D/d)*(600-480)x10⁻⁹ m
=(1.5/2.5*10⁻⁴)*120x10⁻⁹*1000 mm
=(3/5)*12/10 mm
=0.6*1.2 mm
=0.72 mm
1. Find the range of frequency of light that is visible to an average human being (400 nm < 𝜆 < 700 nm).
ANSWER: The velocity of light, c = 3x10⁸ m/s.
Since c = 𝜈𝜆
Hence 𝜈 = c/𝜆
for 𝜆 = 400 nm = 400x10⁻⁹ m
The frequency = 3x10⁸/400x10⁻⁹ Hz =(3/4)x10¹⁵ Hz
= 0.75 x10¹⁵ Hz = 7.5 x 10¹⁴ Hz
For 𝜆 = 700 nm = 700x10⁻⁹ m
The frequency = 3x10⁸/700x10⁻⁹ Hz =(3/7)x10¹⁵ Hz
=0.43x10¹⁵ Hz = 4.3x10¹⁴ Hz.
Hence the range of frequency of light that is visible to an average human being is
4.3x10¹⁴ Hz < 𝜈 <7.5x10¹⁴ Hz.
2. The wavelength of sodium light in air is 589 nm.
(a) Find its frequency in air.
(b) Find its wavelength in water (refractive index = 1.33).
(c) Find its frequency in water.
(d) Find its speed in water.
ANSWER: The wavelength of sodium light in air 𝜆=589 nm =589x10⁻⁹ m. The speed of light in air, c≈3x10⁸ m/s
(a) Hence its frequency in the air, 𝜈 = c/𝜆
=3x10⁸/589x10⁻⁹ Hz
=0.00509x10¹⁷ Hz
=5.09x10¹⁴ Hz
(b) Let the wavelength of sodium light in water =𝜆', and the speed of light in water c'. The frequency does not change, hence
c' =𝜈𝜆', and c = 𝜈𝜆
Thus, c'/c = 𝜆'/𝜆
{but c' = c/µ, where µ is the refractive index of water}
→c/µc = 𝜆'/𝜆
→1/µ = 𝜆'/𝜆
→𝜆' = 𝜆/µ =589/1.33 nm =443 nm
(c) When the light enters another medium, its wavelength changes not the frequency. Hence the frequency in water =𝜈 =5.09x10¹⁴ Hz
(d) The speed of sodium light in water c' = c/µ
=3x10⁸/1.33 m/s
=2.25 x10⁸ m/s
3. The index of refraction of fused quartz is 1.472 for the light of wavelength 400 nm and is 1.452 for the light of wavelength 760 nm. Find the speed of light of these wavelengths in fused quartz.
ANSWER: The speed of light in vacuum, c = 3x10⁸ m/s, The speed of light in the medium c' = c/µ, where µ is the index of refraction. For the light of wavelength 400 nm, given that µ = 1.472. Hence the speed of light for this wavelength
c' = 3x10⁸/1.472 m/s =2.04x10⁸ m/s
For the light of wavelength 750 nm, given that µ = 1.452. Hence the speed of light for this wavelength
c' = 3x10⁸/1.452 m/s =2.07x10⁸ m/s
4. The speed of the yellow light in a certain liquid is 2.4 x 10⁸ m/s. Find the refractive index of the liquid.
ANSWER: Since the speed of light in a medium, c' =c/µ. Where c = speed of light in vacuum and µ =refractive index of the medium. Here given that c' =2.4x10⁸ m/s and we know that c= 3x10⁸ m/s.
Hence, µ = c/c' =3x10⁸/2.4x10⁸ =1.25
5. Two narrow slits emitting light in phase are separated by a distance of 1.0 cm. The wavelength of the light is 5.0 x 10⁻⁷ m. The interference pattern is observed on a screen placed at a distance of 1.0 m.
(a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima?
(b) Find the separation between the sources which will give a separation of 1.0 mm between the consecutive maxima.
ANSWER: The separation of slits, d = 1.0 cm =0.01 m.
The wavelength of light, 𝜆 = 5.0x10⁻⁷ m.
The distance of the screen from the slits, D = 1.0 m.
(a) The separation between consecutive maxima = the fringe width, w = D𝜆/d =1.0*5.0x10⁻⁷/0.01 m
=5x10⁻⁵ *1000 mm =5x10⁻² mm =0.05 mm
The width is in the range of one-hundredth of a millimeter, hence they can hardly be distinguished.
(b) Here given w = 1.0 mm = 0.001 m and d = ?
d = D𝜆/w =1.0*5.0x10⁻⁷/0.001 m
=5x10⁻⁴ *1000 mm =5/10 mm =0.50 mm
6. The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5 m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
ANSWER: Given fringe width, w = 1.0 mm =0.001 m, Distance of the screen, D = 2.5 m, the separation of slits, d = 1.0 mm =0.001 m. The wavelength 𝜆 =?
Since w =D𝜆/d
→𝜆 = wd/D =0.001*0.001/2.5 m =0.4x10⁻⁶ m
=4x10⁻⁷*10⁹ nm =400 nm
7. In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 x 10⁻⁷ m and the distance of the screen from the slits is 1.0 m.
(a) Find the distance of the center of the first minimum from the center of the central maximum.
(b) How many bright fringes are formed in one-centimeter width on the screen?
ANSWER: Given slit separation, d = 1.0 mm =0.001 m, The wavelength of the light, 𝜆 = 5.0x10⁻⁷ m and the distance of the screen, D = 1.0 m.
(a) Hence the fringe width, w =D𝜆/d
→w = 1.0*5.0x10⁻⁷/0.001 m =5x10⁻⁴*1000 mm
→w = 5/10 mm =0.50 mm.
This fringe width is the distance between the center of consecutive maxima. The distance of the center of the first minimum from the center of the central maximum = D𝜆/2d = 0.50/2 mm =0.25 mm
(b) The number of bright fringes in one-centimeter width = 10 mm/0.50 mm = 20
8. In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of the yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?
ANSWER: Given that, the separation of slits, d = 0.800 mm =8x10⁻⁴ m.
The wavelength of the light, 𝜆 = 589 nm =589x10⁻⁹ m.
The screen distance, D = 2.00 m. Fringe width, w =?W =D𝜆/d =2.00*589x10⁻⁹/8x10⁻⁴ m
=147.25*10⁻⁵*1000 mm
≈1.47 mm
9. Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is 2.0 x 10⁻³ m.
ANSWER: Given the slit separation, d = 2.0x10⁻³ m.
The wavelength of the light, 𝜆 =500 nm =500x10⁻⁹ m
Let, w = separation between the consecutive bright fringes.
D = The distance of the screen from the slits and
θ = angular separation between the consecutive bright fringes. Then,
θ = w/D =D𝜆/dD =𝜆/d =500x10⁻⁹/2.0x10⁻³ rad
→θ = 25x10⁻⁵*180°/π =0.014°
10. A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
ANSWER: Given that, the separation between the slits, d = 0.25 mm = 2.5x10⁻⁴ m
The distance of the screen from the slits, D = 150 cm =1.50 m
The first maximum occurs from the central maximum at a distance y =D𝜆/d, where 𝜆 is the wavelength of the light.
For the 𝜆 = 480 nm =480x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y = (D/d)*480x10⁻⁹ m
For the 𝜆' = 600 nm =600x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y' = (D/d)*600x10⁻⁹ m
Hence the linear separation between the first maximum (from the central maximum) corresponding to the two wavelengths =y'-y
=(D/d)*(600-480)x10⁻⁹ m
=(1.5/2.5*10⁻⁴)*120x10⁻⁹*1000 mm
=(3/5)*12/10 mm
=0.6*1.2 mm
=0.72 mm
ANSWER: The velocity of light, c = 3x10⁸ m/s.
Since c = 𝜈𝜆
Hence 𝜈 = c/𝜆
for 𝜆 = 400 nm = 400x10⁻⁹ m
The frequency = 3x10⁸/400x10⁻⁹ Hz =(3/4)x10¹⁵ Hz
= 0.75 x10¹⁵ Hz = 7.5 x 10¹⁴ Hz
For 𝜆 = 700 nm = 700x10⁻⁹ m
The frequency = 3x10⁸/700x10⁻⁹ Hz =(3/7)x10¹⁵ Hz
=0.43x10¹⁵ Hz = 4.3x10¹⁴ Hz.
Hence the range of frequency of light that is visible to an average human being is
4.3x10¹⁴ Hz < 𝜈 <7.5x10¹⁴ Hz.
2. The wavelength of sodium light in air is 589 nm.
(a) Find its frequency in air.
(b) Find its wavelength in water (refractive index = 1.33).
(c) Find its frequency in water.
(d) Find its speed in water.
ANSWER: The wavelength of sodium light in air 𝜆=589 nm =589x10⁻⁹ m. The speed of light in air, c≈3x10⁸ m/s
(a) Hence its frequency in the air, 𝜈 = c/𝜆
=3x10⁸/589x10⁻⁹ Hz
=0.00509x10¹⁷ Hz
=5.09x10¹⁴ Hz
(b) Let the wavelength of sodium light in water =𝜆', and the speed of light in water c'. The frequency does not change, hence
c' =𝜈𝜆', and c = 𝜈𝜆
Thus, c'/c = 𝜆'/𝜆
{but c' = c/µ, where µ is the refractive index of water}
→c/µc = 𝜆'/𝜆
→1/µ = 𝜆'/𝜆
→𝜆' = 𝜆/µ =589/1.33 nm =443 nm
(c) When the light enters another medium, its wavelength changes not the frequency. Hence the frequency in water =𝜈 =5.09x10¹⁴ Hz
(d) The speed of sodium light in water c' = c/µ
=3x10⁸/1.33 m/s
=2.25 x10⁸ m/s
3. The index of refraction of fused quartz is 1.472 for the light of wavelength 400 nm and is 1.452 for the light of wavelength 760 nm. Find the speed of light of these wavelengths in fused quartz.
ANSWER: The speed of light in vacuum, c = 3x10⁸ m/s, The speed of light in the medium c' = c/µ, where µ is the index of refraction. For the light of wavelength 400 nm, given that µ = 1.472. Hence the speed of light for this wavelength
c' = 3x10⁸/1.472 m/s =2.04x10⁸ m/s
For the light of wavelength 750 nm, given that µ = 1.452. Hence the speed of light for this wavelength
c' = 3x10⁸/1.452 m/s =2.07x10⁸ m/s
4. The speed of the yellow light in a certain liquid is 2.4 x 10⁸ m/s. Find the refractive index of the liquid.
ANSWER: Since the speed of light in a medium, c' =c/µ. Where c = speed of light in vacuum and µ =refractive index of the medium. Here given that c' =2.4x10⁸ m/s and we know that c= 3x10⁸ m/s.
Hence, µ = c/c' =3x10⁸/2.4x10⁸ =1.25
5. Two narrow slits emitting light in phase are separated by a distance of 1.0 cm. The wavelength of the light is 5.0 x 10⁻⁷ m. The interference pattern is observed on a screen placed at a distance of 1.0 m.
(a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima?
(b) Find the separation between the sources which will give a separation of 1.0 mm between the consecutive maxima.
ANSWER: The separation of slits, d = 1.0 cm =0.01 m.
The wavelength of light, 𝜆 = 5.0x10⁻⁷ m.
The distance of the screen from the slits, D = 1.0 m.
(a) The separation between consecutive maxima = the fringe width, w = D𝜆/d =1.0*5.0x10⁻⁷/0.01 m
=5x10⁻⁵ *1000 mm =5x10⁻² mm =0.05 mm
The width is in the range of one-hundredth of a millimeter, hence they can hardly be distinguished.
|
| Figure for Q-5 |
(b) Here given w = 1.0 mm = 0.001 m and d = ?
d = D𝜆/w =1.0*5.0x10⁻⁷/0.001 m
=5x10⁻⁴ *1000 mm =5/10 mm =0.50 mm
6. The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5 m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
ANSWER: Given fringe width, w = 1.0 mm =0.001 m, Distance of the screen, D = 2.5 m, the separation of slits, d = 1.0 mm =0.001 m. The wavelength 𝜆 =?
Since w =D𝜆/d
→𝜆 = wd/D =0.001*0.001/2.5 m =0.4x10⁻⁶ m
=4x10⁻⁷*10⁹ nm =400 nm
7. In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 x 10⁻⁷ m and the distance of the screen from the slits is 1.0 m.
(a) Find the distance of the center of the first minimum from the center of the central maximum.
(b) How many bright fringes are formed in one-centimeter width on the screen?
ANSWER: Given slit separation, d = 1.0 mm =0.001 m, The wavelength of the light, 𝜆 = 5.0x10⁻⁷ m and the distance of the screen, D = 1.0 m.
(a) Hence the fringe width, w =D𝜆/d
→w = 1.0*5.0x10⁻⁷/0.001 m =5x10⁻⁴*1000 mm
→w = 5/10 mm =0.50 mm.
This fringe width is the distance between the center of consecutive maxima. The distance of the center of the first minimum from the center of the central maximum = D𝜆/2d = 0.50/2 mm =0.25 mm
(b) The number of bright fringes in one-centimeter width = 10 mm/0.50 mm = 20
8. In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of the yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?
ANSWER: Given that, the separation of slits, d = 0.800 mm =8x10⁻⁴ m.
The wavelength of the light, 𝜆 = 589 nm =589x10⁻⁹ m.
The screen distance, D = 2.00 m. Fringe width, w =?W =D𝜆/d =2.00*589x10⁻⁹/8x10⁻⁴ m
=147.25*10⁻⁵*1000 mm
≈1.47 mm
9. Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is 2.0 x 10⁻³ m.
ANSWER: Given the slit separation, d = 2.0x10⁻³ m.
The wavelength of the light, 𝜆 =500 nm =500x10⁻⁹ m
Let, w = separation between the consecutive bright fringes.
D = The distance of the screen from the slits and
θ = angular separation between the consecutive bright fringes. Then,
θ = w/D =D𝜆/dD =𝜆/d =500x10⁻⁹/2.0x10⁻³ rad
→θ = 25x10⁻⁵*180°/π =0.014°
10. A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
ANSWER: Given that, the separation between the slits, d = 0.25 mm = 2.5x10⁻⁴ m
The distance of the screen from the slits, D = 150 cm =1.50 m
The first maximum occurs from the central maximum at a distance y =D𝜆/d, where 𝜆 is the wavelength of the light.
For the 𝜆 = 480 nm =480x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y = (D/d)*480x10⁻⁹ m
For the 𝜆' = 600 nm =600x10⁻⁹ m, the first maximum is at a distance (from the central maximum)
y' = (D/d)*600x10⁻⁹ m
Hence the linear separation between the first maximum (from the central maximum) corresponding to the two wavelengths =y'-y
=(D/d)*(600-480)x10⁻⁹ m
=(1.5/2.5*10⁻⁴)*120x10⁻⁹*1000 mm
=(3/5)*12/10 mm
=0.6*1.2 mm
=0.72 mm
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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