283303
In Young's double slit experiment, the fringe width is \(\beta\). If the entire arrangement is placed in a liquid of refractive index \(n\), the fringe width becomes:
1 \(\mathrm{n} \beta\)
2 \(\frac{\beta}{\mathrm{n}+1}\)
3 \(\frac{\beta}{\mathrm{n}-1}\)
4 \(\frac{\beta}{\mathrm{n}}\)
Explanation:
: Given, fringe width \(=\beta\), refractive index \(=\mathrm{n}\) We know that, Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) If the entire arrangement is placed in a liquid of refractive index \(n\), then fringe width become \(\beta / n\) i.e. \(\quad \beta^{\prime}=\frac{\beta}{n}\)
Kerala CEE 04.07.2022
WAVE OPTICS
283314
White light is used to illuminate the two slits in a Young's double experiment. The observed central fringe is
1 Black
2 White
3 Blue
4 Red
Explanation:
: White light consists of waves of innumerable wavelengths starting from violet to red colour. 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. Therefore, the central fringe is white.
TS EAMCET 06.08.2021
WAVE OPTICS
283317
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point of the screen corresponding to the fifth minimum has path difference.
1 \(5 \frac{\lambda}{2}\)
2 \(10 \frac{\lambda}{2}\)
3 \(9 \frac{\lambda}{2}\)
4 \(11 \frac{\lambda}{2}\)
Explanation:
: Path difference for \(\mathrm{n}^{\text {th }}\) minima \(\Delta \mathrm{x}_{\mathrm{n}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\) For \(5^{\text {th }}\) minima \(\Delta \mathrm{x}_5=(2 \times 5-1) \frac{\lambda}{2}\) \(=\frac{9 \lambda}{2}\)
NEET (Odisha) - 2020
WAVE OPTICS
283321
A thin transparent sheet of thickness ' \(t\) ' is placed in front of a Young's double slit. The fringe width will be \((\mu>1)\)
1 increase
2 remain same
3 become non-uniform
4 decrease
Explanation:
: Fringe width is given by- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\lambda=\) wavelength \(\mathrm{D}=\) distance of the screen from slit \(\mathrm{d}=\) distance of slits \(\beta\) is independent of thickness, hence no effect on fringe width.
283303
In Young's double slit experiment, the fringe width is \(\beta\). If the entire arrangement is placed in a liquid of refractive index \(n\), the fringe width becomes:
1 \(\mathrm{n} \beta\)
2 \(\frac{\beta}{\mathrm{n}+1}\)
3 \(\frac{\beta}{\mathrm{n}-1}\)
4 \(\frac{\beta}{\mathrm{n}}\)
Explanation:
: Given, fringe width \(=\beta\), refractive index \(=\mathrm{n}\) We know that, Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) If the entire arrangement is placed in a liquid of refractive index \(n\), then fringe width become \(\beta / n\) i.e. \(\quad \beta^{\prime}=\frac{\beta}{n}\)
Kerala CEE 04.07.2022
WAVE OPTICS
283314
White light is used to illuminate the two slits in a Young's double experiment. The observed central fringe is
1 Black
2 White
3 Blue
4 Red
Explanation:
: White light consists of waves of innumerable wavelengths starting from violet to red colour. 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. Therefore, the central fringe is white.
TS EAMCET 06.08.2021
WAVE OPTICS
283317
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point of the screen corresponding to the fifth minimum has path difference.
1 \(5 \frac{\lambda}{2}\)
2 \(10 \frac{\lambda}{2}\)
3 \(9 \frac{\lambda}{2}\)
4 \(11 \frac{\lambda}{2}\)
Explanation:
: Path difference for \(\mathrm{n}^{\text {th }}\) minima \(\Delta \mathrm{x}_{\mathrm{n}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\) For \(5^{\text {th }}\) minima \(\Delta \mathrm{x}_5=(2 \times 5-1) \frac{\lambda}{2}\) \(=\frac{9 \lambda}{2}\)
NEET (Odisha) - 2020
WAVE OPTICS
283321
A thin transparent sheet of thickness ' \(t\) ' is placed in front of a Young's double slit. The fringe width will be \((\mu>1)\)
1 increase
2 remain same
3 become non-uniform
4 decrease
Explanation:
: Fringe width is given by- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\lambda=\) wavelength \(\mathrm{D}=\) distance of the screen from slit \(\mathrm{d}=\) distance of slits \(\beta\) is independent of thickness, hence no effect on fringe width.
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WAVE OPTICS
283303
In Young's double slit experiment, the fringe width is \(\beta\). If the entire arrangement is placed in a liquid of refractive index \(n\), the fringe width becomes:
1 \(\mathrm{n} \beta\)
2 \(\frac{\beta}{\mathrm{n}+1}\)
3 \(\frac{\beta}{\mathrm{n}-1}\)
4 \(\frac{\beta}{\mathrm{n}}\)
Explanation:
: Given, fringe width \(=\beta\), refractive index \(=\mathrm{n}\) We know that, Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) If the entire arrangement is placed in a liquid of refractive index \(n\), then fringe width become \(\beta / n\) i.e. \(\quad \beta^{\prime}=\frac{\beta}{n}\)
Kerala CEE 04.07.2022
WAVE OPTICS
283314
White light is used to illuminate the two slits in a Young's double experiment. The observed central fringe is
1 Black
2 White
3 Blue
4 Red
Explanation:
: White light consists of waves of innumerable wavelengths starting from violet to red colour. 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. Therefore, the central fringe is white.
TS EAMCET 06.08.2021
WAVE OPTICS
283317
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point of the screen corresponding to the fifth minimum has path difference.
1 \(5 \frac{\lambda}{2}\)
2 \(10 \frac{\lambda}{2}\)
3 \(9 \frac{\lambda}{2}\)
4 \(11 \frac{\lambda}{2}\)
Explanation:
: Path difference for \(\mathrm{n}^{\text {th }}\) minima \(\Delta \mathrm{x}_{\mathrm{n}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\) For \(5^{\text {th }}\) minima \(\Delta \mathrm{x}_5=(2 \times 5-1) \frac{\lambda}{2}\) \(=\frac{9 \lambda}{2}\)
NEET (Odisha) - 2020
WAVE OPTICS
283321
A thin transparent sheet of thickness ' \(t\) ' is placed in front of a Young's double slit. The fringe width will be \((\mu>1)\)
1 increase
2 remain same
3 become non-uniform
4 decrease
Explanation:
: Fringe width is given by- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\lambda=\) wavelength \(\mathrm{D}=\) distance of the screen from slit \(\mathrm{d}=\) distance of slits \(\beta\) is independent of thickness, hence no effect on fringe width.
283303
In Young's double slit experiment, the fringe width is \(\beta\). If the entire arrangement is placed in a liquid of refractive index \(n\), the fringe width becomes:
1 \(\mathrm{n} \beta\)
2 \(\frac{\beta}{\mathrm{n}+1}\)
3 \(\frac{\beta}{\mathrm{n}-1}\)
4 \(\frac{\beta}{\mathrm{n}}\)
Explanation:
: Given, fringe width \(=\beta\), refractive index \(=\mathrm{n}\) We know that, Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) If the entire arrangement is placed in a liquid of refractive index \(n\), then fringe width become \(\beta / n\) i.e. \(\quad \beta^{\prime}=\frac{\beta}{n}\)
Kerala CEE 04.07.2022
WAVE OPTICS
283314
White light is used to illuminate the two slits in a Young's double experiment. The observed central fringe is
1 Black
2 White
3 Blue
4 Red
Explanation:
: White light consists of waves of innumerable wavelengths starting from violet to red colour. 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. Therefore, the central fringe is white.
TS EAMCET 06.08.2021
WAVE OPTICS
283317
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point of the screen corresponding to the fifth minimum has path difference.
1 \(5 \frac{\lambda}{2}\)
2 \(10 \frac{\lambda}{2}\)
3 \(9 \frac{\lambda}{2}\)
4 \(11 \frac{\lambda}{2}\)
Explanation:
: Path difference for \(\mathrm{n}^{\text {th }}\) minima \(\Delta \mathrm{x}_{\mathrm{n}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\) For \(5^{\text {th }}\) minima \(\Delta \mathrm{x}_5=(2 \times 5-1) \frac{\lambda}{2}\) \(=\frac{9 \lambda}{2}\)
NEET (Odisha) - 2020
WAVE OPTICS
283321
A thin transparent sheet of thickness ' \(t\) ' is placed in front of a Young's double slit. The fringe width will be \((\mu>1)\)
1 increase
2 remain same
3 become non-uniform
4 decrease
Explanation:
: Fringe width is given by- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\lambda=\) wavelength \(\mathrm{D}=\) distance of the screen from slit \(\mathrm{d}=\) distance of slits \(\beta\) is independent of thickness, hence no effect on fringe width.