NEET Test Series from KOTA - 10 Papers In MS WORD
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283299
In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is
283300
In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.
1 2
2 4
3 8
4 16
Explanation:
: Given that, \(\mathrm{d}_1=\mathrm{d}, \mathrm{d}_2=\frac{\mathrm{d}}{2}, \quad \mathrm{D}_1=\mathrm{D}_2=\mathrm{D}\) \(\lambda_1=\lambda, \lambda_2=4 \lambda\) The position of \(\mathrm{n}^{\text {th }}\) order maxima on the screen is- \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) And, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=\frac{\mathrm{n} \lambda_2 \mathrm{D}_2}{\mathrm{~d}_2}\) \(x_n^{\prime}=\frac{n 4 \lambda \times D}{d / 2}\) \(x_n^{\prime}=8\left(\frac{n \lambda D}{d}\right)\)So, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=8 \mathrm{x}_{\mathrm{n}}\)
TS EAMCET 20.07.2022
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283302
In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?
1 \(1 / 9\)
2 \(1 / 3\)
3 \(1 / 16\)
4 \(1 / 2\)
Explanation:
: Given that, \(\mathrm{I}_{\text {min }}=\mathrm{I}_0\) \(\mathrm{I}_{\max }=4 \mathrm{I}_0\) We know that, \(\mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) \(4 \mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) And, \(\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) \(\mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) Form equation (i) and equation (ii), we get- \(\frac{4 \mathrm{I}_0}{\mathrm{I}_0}=\left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2\) \(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}=2\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)\) \(\sqrt{\mathrm{I}_1}=3 \sqrt{\mathrm{I}_2}\)Hence, \(\quad \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{9}\)
WB JEE 2022
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283304
In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(I, \lambda\) being the wavelength of light used. The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be
1 \(\frac{I}{4}\)
2 \(\frac{I}{2}\)
3 I
4 zero
Explanation:
: We know that, Phase difference \((\phi)=\left(\frac{2 \pi}{\lambda}\right) \times\) path difference \(=\left(\frac{2 \pi}{\lambda}\right) \Delta \mathrm{x}\) Case (i) for path difference \(\lambda\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \mathrm{x} \lambda=2 \pi\) Case (ii) for path difference \(\lambda / 4\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \times \frac{\lambda}{4}=\frac{\pi}{2}\) Intensity on screen, \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right)\) For \(\phi=2 \pi\) Then, \(\quad I=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)\) Therefore, \(\mathrm{I}=4 \mathrm{I}_0\) For \(\phi=\left(\frac{\pi}{2}\right)\) then, \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)\) \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \times \frac{1}{2}\) \(\mathrm{I}^{\prime}=\frac{\mathrm{I}}{2} \quad\left(\because \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{I}}{4}\right)\)
283299
In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is
283300
In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.
1 2
2 4
3 8
4 16
Explanation:
: Given that, \(\mathrm{d}_1=\mathrm{d}, \mathrm{d}_2=\frac{\mathrm{d}}{2}, \quad \mathrm{D}_1=\mathrm{D}_2=\mathrm{D}\) \(\lambda_1=\lambda, \lambda_2=4 \lambda\) The position of \(\mathrm{n}^{\text {th }}\) order maxima on the screen is- \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) And, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=\frac{\mathrm{n} \lambda_2 \mathrm{D}_2}{\mathrm{~d}_2}\) \(x_n^{\prime}=\frac{n 4 \lambda \times D}{d / 2}\) \(x_n^{\prime}=8\left(\frac{n \lambda D}{d}\right)\)So, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=8 \mathrm{x}_{\mathrm{n}}\)
TS EAMCET 20.07.2022
WAVE OPTICS
283302
In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?
1 \(1 / 9\)
2 \(1 / 3\)
3 \(1 / 16\)
4 \(1 / 2\)
Explanation:
: Given that, \(\mathrm{I}_{\text {min }}=\mathrm{I}_0\) \(\mathrm{I}_{\max }=4 \mathrm{I}_0\) We know that, \(\mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) \(4 \mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) And, \(\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) \(\mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) Form equation (i) and equation (ii), we get- \(\frac{4 \mathrm{I}_0}{\mathrm{I}_0}=\left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2\) \(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}=2\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)\) \(\sqrt{\mathrm{I}_1}=3 \sqrt{\mathrm{I}_2}\)Hence, \(\quad \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{9}\)
WB JEE 2022
WAVE OPTICS
283304
In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(I, \lambda\) being the wavelength of light used. The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be
1 \(\frac{I}{4}\)
2 \(\frac{I}{2}\)
3 I
4 zero
Explanation:
: We know that, Phase difference \((\phi)=\left(\frac{2 \pi}{\lambda}\right) \times\) path difference \(=\left(\frac{2 \pi}{\lambda}\right) \Delta \mathrm{x}\) Case (i) for path difference \(\lambda\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \mathrm{x} \lambda=2 \pi\) Case (ii) for path difference \(\lambda / 4\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \times \frac{\lambda}{4}=\frac{\pi}{2}\) Intensity on screen, \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right)\) For \(\phi=2 \pi\) Then, \(\quad I=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)\) Therefore, \(\mathrm{I}=4 \mathrm{I}_0\) For \(\phi=\left(\frac{\pi}{2}\right)\) then, \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)\) \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \times \frac{1}{2}\) \(\mathrm{I}^{\prime}=\frac{\mathrm{I}}{2} \quad\left(\because \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{I}}{4}\right)\)
283299
In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is
283300
In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.
1 2
2 4
3 8
4 16
Explanation:
: Given that, \(\mathrm{d}_1=\mathrm{d}, \mathrm{d}_2=\frac{\mathrm{d}}{2}, \quad \mathrm{D}_1=\mathrm{D}_2=\mathrm{D}\) \(\lambda_1=\lambda, \lambda_2=4 \lambda\) The position of \(\mathrm{n}^{\text {th }}\) order maxima on the screen is- \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) And, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=\frac{\mathrm{n} \lambda_2 \mathrm{D}_2}{\mathrm{~d}_2}\) \(x_n^{\prime}=\frac{n 4 \lambda \times D}{d / 2}\) \(x_n^{\prime}=8\left(\frac{n \lambda D}{d}\right)\)So, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=8 \mathrm{x}_{\mathrm{n}}\)
TS EAMCET 20.07.2022
WAVE OPTICS
283302
In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?
1 \(1 / 9\)
2 \(1 / 3\)
3 \(1 / 16\)
4 \(1 / 2\)
Explanation:
: Given that, \(\mathrm{I}_{\text {min }}=\mathrm{I}_0\) \(\mathrm{I}_{\max }=4 \mathrm{I}_0\) We know that, \(\mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) \(4 \mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) And, \(\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) \(\mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) Form equation (i) and equation (ii), we get- \(\frac{4 \mathrm{I}_0}{\mathrm{I}_0}=\left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2\) \(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}=2\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)\) \(\sqrt{\mathrm{I}_1}=3 \sqrt{\mathrm{I}_2}\)Hence, \(\quad \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{9}\)
WB JEE 2022
WAVE OPTICS
283304
In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(I, \lambda\) being the wavelength of light used. The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be
1 \(\frac{I}{4}\)
2 \(\frac{I}{2}\)
3 I
4 zero
Explanation:
: We know that, Phase difference \((\phi)=\left(\frac{2 \pi}{\lambda}\right) \times\) path difference \(=\left(\frac{2 \pi}{\lambda}\right) \Delta \mathrm{x}\) Case (i) for path difference \(\lambda\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \mathrm{x} \lambda=2 \pi\) Case (ii) for path difference \(\lambda / 4\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \times \frac{\lambda}{4}=\frac{\pi}{2}\) Intensity on screen, \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right)\) For \(\phi=2 \pi\) Then, \(\quad I=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)\) Therefore, \(\mathrm{I}=4 \mathrm{I}_0\) For \(\phi=\left(\frac{\pi}{2}\right)\) then, \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)\) \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \times \frac{1}{2}\) \(\mathrm{I}^{\prime}=\frac{\mathrm{I}}{2} \quad\left(\because \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{I}}{4}\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
WAVE OPTICS
283299
In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is
283300
In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.
1 2
2 4
3 8
4 16
Explanation:
: Given that, \(\mathrm{d}_1=\mathrm{d}, \mathrm{d}_2=\frac{\mathrm{d}}{2}, \quad \mathrm{D}_1=\mathrm{D}_2=\mathrm{D}\) \(\lambda_1=\lambda, \lambda_2=4 \lambda\) The position of \(\mathrm{n}^{\text {th }}\) order maxima on the screen is- \(\mathrm{x}_{\mathrm{n}}=\frac{\mathrm{n} \lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) And, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=\frac{\mathrm{n} \lambda_2 \mathrm{D}_2}{\mathrm{~d}_2}\) \(x_n^{\prime}=\frac{n 4 \lambda \times D}{d / 2}\) \(x_n^{\prime}=8\left(\frac{n \lambda D}{d}\right)\)So, \(\quad \mathrm{x}_{\mathrm{n}}^{\prime}=8 \mathrm{x}_{\mathrm{n}}\)
TS EAMCET 20.07.2022
WAVE OPTICS
283302
In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?
1 \(1 / 9\)
2 \(1 / 3\)
3 \(1 / 16\)
4 \(1 / 2\)
Explanation:
: Given that, \(\mathrm{I}_{\text {min }}=\mathrm{I}_0\) \(\mathrm{I}_{\max }=4 \mathrm{I}_0\) We know that, \(\mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) \(4 \mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) And, \(\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) \(\mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\) Form equation (i) and equation (ii), we get- \(\frac{4 \mathrm{I}_0}{\mathrm{I}_0}=\left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2\) \(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}=2\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)\) \(\sqrt{\mathrm{I}_1}=3 \sqrt{\mathrm{I}_2}\)Hence, \(\quad \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{9}\)
WB JEE 2022
WAVE OPTICS
283304
In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(I, \lambda\) being the wavelength of light used. The intensity at a point where the path difference is \(\frac{\lambda}{4}\) will be
1 \(\frac{I}{4}\)
2 \(\frac{I}{2}\)
3 I
4 zero
Explanation:
: We know that, Phase difference \((\phi)=\left(\frac{2 \pi}{\lambda}\right) \times\) path difference \(=\left(\frac{2 \pi}{\lambda}\right) \Delta \mathrm{x}\) Case (i) for path difference \(\lambda\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \mathrm{x} \lambda=2 \pi\) Case (ii) for path difference \(\lambda / 4\) \(\phi=\left(\frac{2 \pi}{\lambda}\right) \times \frac{\lambda}{4}=\frac{\pi}{2}\) Intensity on screen, \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right)\) For \(\phi=2 \pi\) Then, \(\quad I=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)\) Therefore, \(\mathrm{I}=4 \mathrm{I}_0\) For \(\phi=\left(\frac{\pi}{2}\right)\) then, \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)\) \(\mathrm{I}^{\prime}=4 \mathrm{I}_0 \times \frac{1}{2}\) \(\mathrm{I}^{\prime}=\frac{\mathrm{I}}{2} \quad\left(\because \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{I}}{4}\right)\)