283350
In Young's double slit experiment, \(\lambda=500 \mathrm{~nm}\), \(d=1 \mathrm{~mm}, D=1 \mathrm{~m}\). Minimum distance from the central maximum for which intensity is half of the maximum intensity is
1 \(2.5 \times 10^{-4} \mathrm{~m}\)
2 \(1.25 \times 10^{-4} \mathrm{~m}\)
3 \(0.625 \times 10^{-4} \mathrm{~m}\)
4 \(0.3125 \times 10^{-4}\)
Explanation:
: Given, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{~d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\) \(\mathrm{D}=1 \mathrm{~m}\) Let, \(y\) be the minimum distance for intensity is half of the maximum intensity then, \(\frac{I_o}{2}=I_o \cos ^2\left(\frac{\pi y}{\beta}\right)\) \(\frac{\pi y}{\beta}=\frac{\pi}{4}\) \(y=\frac{\lambda D}{4 d} \quad\left(\because \beta=\frac{D \lambda}{d}\right)\) \(y=\frac{500 \times 10^{-9} \times 1}{4 \times 10^{-3}}\) \(y=1.25 \times 10^{-4} \mathrm{~m}\)
BITSAT-2018
WAVE OPTICS
283351
In Young's experiment, using red light \((\lambda=6600 \AA), 60\) fringes are seen in the field of view. How many fringes will be seen by using violet light \((\lambda=4400 \AA)\) ?
1 10
2 20
3 45
4 90
Explanation:
: Given, Wavelength of red light \(\left(\lambda_1\right)=6600 \AA\) No. of fringes \(\left(\mathrm{n}_1\right)=60\) fringes Wavelength of violet light \(\left(\lambda_2\right)=4400 \AA\) No of fringes \(\left(\mathrm{n}_2\right)=\) ? The number of fringes is given by \(\mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2\) \(60 \times 6600=\mathrm{n}_2 \times 4400\) \(\mathrm{n}_2=\frac{60 \times 6600}{4400}\) \(=90 \text { fringes. }\)Hence, 90 fringes will be seen by using violet light.
TS- EAMCET-10.09.2020
WAVE OPTICS
283352
Light of wavelength \(5500 \AA\) from narrow slit is incident on a double slit. The overall separation of 5 fringes on a screen \(200 \mathrm{~cm}\) away is \(1 \mathrm{~cm}\). Calculate slit separation.
283353
Young's double slit experiments is first performed in air and then in a medium other than air. It is found that \(8^{\text {th }}\) bright fringe in the medium lies where \(5^{\text {th }}\) dark fringe lies in air. The refractive index of the medium is nearly
1 1.25
2 1.59
3 1.69
4 1.78
Explanation:
: The position of \(\mathrm{n}^{\text {th }}\) bright fringe \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) \(8^{\text {th }}\) bright fringe in medium \(\mathrm{x}_8=\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}\) The position of \(\mathrm{n}^{\text {th }}\) dark fringe \(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) \(5^{\text {th }}\) dark fringe in air \(x_5=\frac{(2 \times 5-1) \lambda D}{2 d}\) \(x_5=\frac{9 \lambda_{\text {air }} D}{2 d}\) Given, \(\left(\mathrm{x}_8\right)_{\text {bright }}=\left(\mathrm{x}_5\right)_{\text {dark }}\) \(\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}=\frac{9 \lambda_{\text {air }} \mathrm{D}}{2 \mathrm{~d}}\) \(\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\frac{\mu_{\text {med }}}{\mu_{\text {air }}}=\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\mu_{\text {med }}=1.78 \mu_{\text {air }} \quad\left(\mu_{\text {air }}=1\right)\)
283350
In Young's double slit experiment, \(\lambda=500 \mathrm{~nm}\), \(d=1 \mathrm{~mm}, D=1 \mathrm{~m}\). Minimum distance from the central maximum for which intensity is half of the maximum intensity is
1 \(2.5 \times 10^{-4} \mathrm{~m}\)
2 \(1.25 \times 10^{-4} \mathrm{~m}\)
3 \(0.625 \times 10^{-4} \mathrm{~m}\)
4 \(0.3125 \times 10^{-4}\)
Explanation:
: Given, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{~d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\) \(\mathrm{D}=1 \mathrm{~m}\) Let, \(y\) be the minimum distance for intensity is half of the maximum intensity then, \(\frac{I_o}{2}=I_o \cos ^2\left(\frac{\pi y}{\beta}\right)\) \(\frac{\pi y}{\beta}=\frac{\pi}{4}\) \(y=\frac{\lambda D}{4 d} \quad\left(\because \beta=\frac{D \lambda}{d}\right)\) \(y=\frac{500 \times 10^{-9} \times 1}{4 \times 10^{-3}}\) \(y=1.25 \times 10^{-4} \mathrm{~m}\)
BITSAT-2018
WAVE OPTICS
283351
In Young's experiment, using red light \((\lambda=6600 \AA), 60\) fringes are seen in the field of view. How many fringes will be seen by using violet light \((\lambda=4400 \AA)\) ?
1 10
2 20
3 45
4 90
Explanation:
: Given, Wavelength of red light \(\left(\lambda_1\right)=6600 \AA\) No. of fringes \(\left(\mathrm{n}_1\right)=60\) fringes Wavelength of violet light \(\left(\lambda_2\right)=4400 \AA\) No of fringes \(\left(\mathrm{n}_2\right)=\) ? The number of fringes is given by \(\mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2\) \(60 \times 6600=\mathrm{n}_2 \times 4400\) \(\mathrm{n}_2=\frac{60 \times 6600}{4400}\) \(=90 \text { fringes. }\)Hence, 90 fringes will be seen by using violet light.
TS- EAMCET-10.09.2020
WAVE OPTICS
283352
Light of wavelength \(5500 \AA\) from narrow slit is incident on a double slit. The overall separation of 5 fringes on a screen \(200 \mathrm{~cm}\) away is \(1 \mathrm{~cm}\). Calculate slit separation.
283353
Young's double slit experiments is first performed in air and then in a medium other than air. It is found that \(8^{\text {th }}\) bright fringe in the medium lies where \(5^{\text {th }}\) dark fringe lies in air. The refractive index of the medium is nearly
1 1.25
2 1.59
3 1.69
4 1.78
Explanation:
: The position of \(\mathrm{n}^{\text {th }}\) bright fringe \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) \(8^{\text {th }}\) bright fringe in medium \(\mathrm{x}_8=\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}\) The position of \(\mathrm{n}^{\text {th }}\) dark fringe \(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) \(5^{\text {th }}\) dark fringe in air \(x_5=\frac{(2 \times 5-1) \lambda D}{2 d}\) \(x_5=\frac{9 \lambda_{\text {air }} D}{2 d}\) Given, \(\left(\mathrm{x}_8\right)_{\text {bright }}=\left(\mathrm{x}_5\right)_{\text {dark }}\) \(\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}=\frac{9 \lambda_{\text {air }} \mathrm{D}}{2 \mathrm{~d}}\) \(\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\frac{\mu_{\text {med }}}{\mu_{\text {air }}}=\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\mu_{\text {med }}=1.78 \mu_{\text {air }} \quad\left(\mu_{\text {air }}=1\right)\)
283350
In Young's double slit experiment, \(\lambda=500 \mathrm{~nm}\), \(d=1 \mathrm{~mm}, D=1 \mathrm{~m}\). Minimum distance from the central maximum for which intensity is half of the maximum intensity is
1 \(2.5 \times 10^{-4} \mathrm{~m}\)
2 \(1.25 \times 10^{-4} \mathrm{~m}\)
3 \(0.625 \times 10^{-4} \mathrm{~m}\)
4 \(0.3125 \times 10^{-4}\)
Explanation:
: Given, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{~d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\) \(\mathrm{D}=1 \mathrm{~m}\) Let, \(y\) be the minimum distance for intensity is half of the maximum intensity then, \(\frac{I_o}{2}=I_o \cos ^2\left(\frac{\pi y}{\beta}\right)\) \(\frac{\pi y}{\beta}=\frac{\pi}{4}\) \(y=\frac{\lambda D}{4 d} \quad\left(\because \beta=\frac{D \lambda}{d}\right)\) \(y=\frac{500 \times 10^{-9} \times 1}{4 \times 10^{-3}}\) \(y=1.25 \times 10^{-4} \mathrm{~m}\)
BITSAT-2018
WAVE OPTICS
283351
In Young's experiment, using red light \((\lambda=6600 \AA), 60\) fringes are seen in the field of view. How many fringes will be seen by using violet light \((\lambda=4400 \AA)\) ?
1 10
2 20
3 45
4 90
Explanation:
: Given, Wavelength of red light \(\left(\lambda_1\right)=6600 \AA\) No. of fringes \(\left(\mathrm{n}_1\right)=60\) fringes Wavelength of violet light \(\left(\lambda_2\right)=4400 \AA\) No of fringes \(\left(\mathrm{n}_2\right)=\) ? The number of fringes is given by \(\mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2\) \(60 \times 6600=\mathrm{n}_2 \times 4400\) \(\mathrm{n}_2=\frac{60 \times 6600}{4400}\) \(=90 \text { fringes. }\)Hence, 90 fringes will be seen by using violet light.
TS- EAMCET-10.09.2020
WAVE OPTICS
283352
Light of wavelength \(5500 \AA\) from narrow slit is incident on a double slit. The overall separation of 5 fringes on a screen \(200 \mathrm{~cm}\) away is \(1 \mathrm{~cm}\). Calculate slit separation.
283353
Young's double slit experiments is first performed in air and then in a medium other than air. It is found that \(8^{\text {th }}\) bright fringe in the medium lies where \(5^{\text {th }}\) dark fringe lies in air. The refractive index of the medium is nearly
1 1.25
2 1.59
3 1.69
4 1.78
Explanation:
: The position of \(\mathrm{n}^{\text {th }}\) bright fringe \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) \(8^{\text {th }}\) bright fringe in medium \(\mathrm{x}_8=\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}\) The position of \(\mathrm{n}^{\text {th }}\) dark fringe \(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) \(5^{\text {th }}\) dark fringe in air \(x_5=\frac{(2 \times 5-1) \lambda D}{2 d}\) \(x_5=\frac{9 \lambda_{\text {air }} D}{2 d}\) Given, \(\left(\mathrm{x}_8\right)_{\text {bright }}=\left(\mathrm{x}_5\right)_{\text {dark }}\) \(\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}=\frac{9 \lambda_{\text {air }} \mathrm{D}}{2 \mathrm{~d}}\) \(\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\frac{\mu_{\text {med }}}{\mu_{\text {air }}}=\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\mu_{\text {med }}=1.78 \mu_{\text {air }} \quad\left(\mu_{\text {air }}=1\right)\)
283350
In Young's double slit experiment, \(\lambda=500 \mathrm{~nm}\), \(d=1 \mathrm{~mm}, D=1 \mathrm{~m}\). Minimum distance from the central maximum for which intensity is half of the maximum intensity is
1 \(2.5 \times 10^{-4} \mathrm{~m}\)
2 \(1.25 \times 10^{-4} \mathrm{~m}\)
3 \(0.625 \times 10^{-4} \mathrm{~m}\)
4 \(0.3125 \times 10^{-4}\)
Explanation:
: Given, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{~d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\) \(\mathrm{D}=1 \mathrm{~m}\) Let, \(y\) be the minimum distance for intensity is half of the maximum intensity then, \(\frac{I_o}{2}=I_o \cos ^2\left(\frac{\pi y}{\beta}\right)\) \(\frac{\pi y}{\beta}=\frac{\pi}{4}\) \(y=\frac{\lambda D}{4 d} \quad\left(\because \beta=\frac{D \lambda}{d}\right)\) \(y=\frac{500 \times 10^{-9} \times 1}{4 \times 10^{-3}}\) \(y=1.25 \times 10^{-4} \mathrm{~m}\)
BITSAT-2018
WAVE OPTICS
283351
In Young's experiment, using red light \((\lambda=6600 \AA), 60\) fringes are seen in the field of view. How many fringes will be seen by using violet light \((\lambda=4400 \AA)\) ?
1 10
2 20
3 45
4 90
Explanation:
: Given, Wavelength of red light \(\left(\lambda_1\right)=6600 \AA\) No. of fringes \(\left(\mathrm{n}_1\right)=60\) fringes Wavelength of violet light \(\left(\lambda_2\right)=4400 \AA\) No of fringes \(\left(\mathrm{n}_2\right)=\) ? The number of fringes is given by \(\mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2\) \(60 \times 6600=\mathrm{n}_2 \times 4400\) \(\mathrm{n}_2=\frac{60 \times 6600}{4400}\) \(=90 \text { fringes. }\)Hence, 90 fringes will be seen by using violet light.
TS- EAMCET-10.09.2020
WAVE OPTICS
283352
Light of wavelength \(5500 \AA\) from narrow slit is incident on a double slit. The overall separation of 5 fringes on a screen \(200 \mathrm{~cm}\) away is \(1 \mathrm{~cm}\). Calculate slit separation.
283353
Young's double slit experiments is first performed in air and then in a medium other than air. It is found that \(8^{\text {th }}\) bright fringe in the medium lies where \(5^{\text {th }}\) dark fringe lies in air. The refractive index of the medium is nearly
1 1.25
2 1.59
3 1.69
4 1.78
Explanation:
: The position of \(\mathrm{n}^{\text {th }}\) bright fringe \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) \(8^{\text {th }}\) bright fringe in medium \(\mathrm{x}_8=\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}\) The position of \(\mathrm{n}^{\text {th }}\) dark fringe \(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) \(5^{\text {th }}\) dark fringe in air \(x_5=\frac{(2 \times 5-1) \lambda D}{2 d}\) \(x_5=\frac{9 \lambda_{\text {air }} D}{2 d}\) Given, \(\left(\mathrm{x}_8\right)_{\text {bright }}=\left(\mathrm{x}_5\right)_{\text {dark }}\) \(\frac{8 \lambda_{\text {med }} \mathrm{D}}{\mathrm{d}}=\frac{9 \lambda_{\text {air }} \mathrm{D}}{2 \mathrm{~d}}\) \(\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\frac{\mu_{\text {med }}}{\mu_{\text {air }}}=\frac{\lambda_{\text {air }}}{\lambda_{\text {med }}}=\frac{16}{9}\) \(\mu_{\text {med }}=1.78 \mu_{\text {air }} \quad\left(\mu_{\text {air }}=1\right)\)
