283344
In a Young's double slit experiment, the spacing between the two slits is \(0.1 \mathrm{~mm}\). If the screen is kept at a distance of \(1.0 \mathrm{~m}\) from the slit and the wavelength of light is \(5000 \AA\), then the fringe width will be
283153
Young's double slit experiment is performed in water, instead of air, then fringe width
1 becomes infinite
2 remains same
3 decreases
4 increases
Explanation:
: In water speed of light decrease and frequency remains same. \(\mathrm{v}=\mathrm{f} \lambda\) If \(\mathrm{V}\) decrease then \(\lambda\) will also decrease fringe width \((\beta)\) \(=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\mathrm{D}\) and \(\mathrm{d}\) are constant If \(\lambda\) decreases the \(\beta\) will also decrease.
MHT-CET 2020
WAVE OPTICS
283158
In Young's double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Using red light instead of green light
2 Moving the screen away from the slits
3 Reducing the separation between the slits
4 Using blue light instead of green light
Explanation:
: For fringes to be more closely spaced fringe width \((\beta)\) must be decreased. For this wavelength must decrease. Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Hence, blue light has low wavelength as compare to green light.
MHT-CET 2020
WAVE OPTICS
283160
When two coherent sources in Young's experiment are far apart, then interference pattern
1 is not detected.
2 is completely dark.
3 consists of widely separated bright fringes.
4 will be sharp and clear.
Explanation:
: When source are far apart- Fringe width \((\beta)=\frac{D \lambda}{d}\) Then, \(d\) will be large and \(\beta\) will be so small, fringes do not appear separate and interference pattern is not detected.
283344
In a Young's double slit experiment, the spacing between the two slits is \(0.1 \mathrm{~mm}\). If the screen is kept at a distance of \(1.0 \mathrm{~m}\) from the slit and the wavelength of light is \(5000 \AA\), then the fringe width will be
283153
Young's double slit experiment is performed in water, instead of air, then fringe width
1 becomes infinite
2 remains same
3 decreases
4 increases
Explanation:
: In water speed of light decrease and frequency remains same. \(\mathrm{v}=\mathrm{f} \lambda\) If \(\mathrm{V}\) decrease then \(\lambda\) will also decrease fringe width \((\beta)\) \(=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\mathrm{D}\) and \(\mathrm{d}\) are constant If \(\lambda\) decreases the \(\beta\) will also decrease.
MHT-CET 2020
WAVE OPTICS
283158
In Young's double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Using red light instead of green light
2 Moving the screen away from the slits
3 Reducing the separation between the slits
4 Using blue light instead of green light
Explanation:
: For fringes to be more closely spaced fringe width \((\beta)\) must be decreased. For this wavelength must decrease. Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Hence, blue light has low wavelength as compare to green light.
MHT-CET 2020
WAVE OPTICS
283160
When two coherent sources in Young's experiment are far apart, then interference pattern
1 is not detected.
2 is completely dark.
3 consists of widely separated bright fringes.
4 will be sharp and clear.
Explanation:
: When source are far apart- Fringe width \((\beta)=\frac{D \lambda}{d}\) Then, \(d\) will be large and \(\beta\) will be so small, fringes do not appear separate and interference pattern is not detected.
283344
In a Young's double slit experiment, the spacing between the two slits is \(0.1 \mathrm{~mm}\). If the screen is kept at a distance of \(1.0 \mathrm{~m}\) from the slit and the wavelength of light is \(5000 \AA\), then the fringe width will be
283153
Young's double slit experiment is performed in water, instead of air, then fringe width
1 becomes infinite
2 remains same
3 decreases
4 increases
Explanation:
: In water speed of light decrease and frequency remains same. \(\mathrm{v}=\mathrm{f} \lambda\) If \(\mathrm{V}\) decrease then \(\lambda\) will also decrease fringe width \((\beta)\) \(=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\mathrm{D}\) and \(\mathrm{d}\) are constant If \(\lambda\) decreases the \(\beta\) will also decrease.
MHT-CET 2020
WAVE OPTICS
283158
In Young's double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Using red light instead of green light
2 Moving the screen away from the slits
3 Reducing the separation between the slits
4 Using blue light instead of green light
Explanation:
: For fringes to be more closely spaced fringe width \((\beta)\) must be decreased. For this wavelength must decrease. Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Hence, blue light has low wavelength as compare to green light.
MHT-CET 2020
WAVE OPTICS
283160
When two coherent sources in Young's experiment are far apart, then interference pattern
1 is not detected.
2 is completely dark.
3 consists of widely separated bright fringes.
4 will be sharp and clear.
Explanation:
: When source are far apart- Fringe width \((\beta)=\frac{D \lambda}{d}\) Then, \(d\) will be large and \(\beta\) will be so small, fringes do not appear separate and interference pattern is not detected.
283344
In a Young's double slit experiment, the spacing between the two slits is \(0.1 \mathrm{~mm}\). If the screen is kept at a distance of \(1.0 \mathrm{~m}\) from the slit and the wavelength of light is \(5000 \AA\), then the fringe width will be
283153
Young's double slit experiment is performed in water, instead of air, then fringe width
1 becomes infinite
2 remains same
3 decreases
4 increases
Explanation:
: In water speed of light decrease and frequency remains same. \(\mathrm{v}=\mathrm{f} \lambda\) If \(\mathrm{V}\) decrease then \(\lambda\) will also decrease fringe width \((\beta)\) \(=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\mathrm{D}\) and \(\mathrm{d}\) are constant If \(\lambda\) decreases the \(\beta\) will also decrease.
MHT-CET 2020
WAVE OPTICS
283158
In Young's double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Using red light instead of green light
2 Moving the screen away from the slits
3 Reducing the separation between the slits
4 Using blue light instead of green light
Explanation:
: For fringes to be more closely spaced fringe width \((\beta)\) must be decreased. For this wavelength must decrease. Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Hence, blue light has low wavelength as compare to green light.
MHT-CET 2020
WAVE OPTICS
283160
When two coherent sources in Young's experiment are far apart, then interference pattern
1 is not detected.
2 is completely dark.
3 consists of widely separated bright fringes.
4 will be sharp and clear.
Explanation:
: When source are far apart- Fringe width \((\beta)=\frac{D \lambda}{d}\) Then, \(d\) will be large and \(\beta\) will be so small, fringes do not appear separate and interference pattern is not detected.
