283205
Red light of wavelength \(625 \mathrm{~nm}\) is incident normally on an optical diffraction grating with \(2 \times 10^5\) lines \(/ \mathrm{m}\). Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?
1 15
2 17
3 20
4 18
Explanation:
: We know that, For principle maxima in grating spectra - \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=1,2,3\) is the order of principle maximum and \(\theta\) is the angle of diffraction. So, \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 10^5}=8\) Now, number of maxima \(=(2 n+1)\) \(=(2 \times 8+1)=17\)
UPSEE - 2014
WAVE OPTICS
283206
A double slit experiment is performed with light of wavelength \(500 \mathrm{~nm}\). A thin film of thickness \(2 \mu \mathrm{m}\) and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
1 remain unshifted
2 shift downward by nearly two fringes
3 shift upward by nearly two fringes
4 shift downward by its fringes
Explanation:
: Given that, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{t}=2 \mu \mathrm{m}=2 \times 10^{-6} \mathrm{~m}\) Refractive index \((\mu)=1.5\) \(\mathrm{t}=2 \mu \mathrm{m}\) Extra path difference \(=(\mu-1) t\) Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\beta\) is unchanged, path difference will be shifted. So, \(\quad(\mu-1) \mathrm{t}=\mathrm{n} \lambda\) \(\mathrm{n}=\frac{(\mu-1) \mathrm{t}}{\lambda}=\frac{(1.5-1) \times 2 \times 10^{-6}}{500 \times 10^{-9}}=2\)Light will be shifted upward by nearly two fringes.
UPSEE - 2014
WAVE OPTICS
283208
A slit of width a is illuminated by white light. For red light \((\lambda=6200 \AA)\), the first minima is obtained at a diffraction angle of \(30^{\circ}\). Then the value of \(a\) is
1 \(3250 \AA\)
2 \(6.5 \times 10^{-4} \mathrm{~mm}\)
3 1.24 micron
4 \(2.6 \times 10^{-4} \mathrm{~cm}\)
Explanation:
: Given that, \(\lambda=6200 \AA=6200 \times 10^{-10} \mathrm{~m}, \theta=\) \(30^{\circ}\) We know that, For first minima- \(\mathrm{a} \sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}=\frac{6200 \times 10^{-10}}{\sin 30^{\circ}}\) \(\mathrm{a}=1.24 \times 10^{-6} \mathrm{~m}\) \(\mathrm{a}=1.24 \mu\)
UPSEE - 2010
WAVE OPTICS
283212
In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda=\) wavelength of the light \()\) is \(I\). If \(I_0\) denotes the maximum intensity, then \(\frac{I}{I_0}\) is equal to
1 \(\frac{1}{2}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{3}{4}\)
5 \(\frac{1}{\sqrt{3}}\)
Explanation:
: Given, \(\Delta x=\frac{\lambda}{6}\) We know, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\) \(\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}\) \(\Delta \phi=\frac{\pi}{3}\) Now, \(\quad I^{\prime}=I_1+I_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi\) And, \(\quad I^{\prime}=I+I+2 I \cos \frac{\pi}{3}=3 I\) And, \(\quad \mathrm{I}_{\mathrm{o}}=\mathrm{I}+\mathrm{I}+2 \mathrm{I} \cos 0^{\circ}=4 \mathrm{I}\) From equation (i) and (ii), we get- \(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}=\frac{3}{4}\)
283205
Red light of wavelength \(625 \mathrm{~nm}\) is incident normally on an optical diffraction grating with \(2 \times 10^5\) lines \(/ \mathrm{m}\). Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?
1 15
2 17
3 20
4 18
Explanation:
: We know that, For principle maxima in grating spectra - \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=1,2,3\) is the order of principle maximum and \(\theta\) is the angle of diffraction. So, \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 10^5}=8\) Now, number of maxima \(=(2 n+1)\) \(=(2 \times 8+1)=17\)
UPSEE - 2014
WAVE OPTICS
283206
A double slit experiment is performed with light of wavelength \(500 \mathrm{~nm}\). A thin film of thickness \(2 \mu \mathrm{m}\) and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
1 remain unshifted
2 shift downward by nearly two fringes
3 shift upward by nearly two fringes
4 shift downward by its fringes
Explanation:
: Given that, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{t}=2 \mu \mathrm{m}=2 \times 10^{-6} \mathrm{~m}\) Refractive index \((\mu)=1.5\) \(\mathrm{t}=2 \mu \mathrm{m}\) Extra path difference \(=(\mu-1) t\) Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\beta\) is unchanged, path difference will be shifted. So, \(\quad(\mu-1) \mathrm{t}=\mathrm{n} \lambda\) \(\mathrm{n}=\frac{(\mu-1) \mathrm{t}}{\lambda}=\frac{(1.5-1) \times 2 \times 10^{-6}}{500 \times 10^{-9}}=2\)Light will be shifted upward by nearly two fringes.
UPSEE - 2014
WAVE OPTICS
283208
A slit of width a is illuminated by white light. For red light \((\lambda=6200 \AA)\), the first minima is obtained at a diffraction angle of \(30^{\circ}\). Then the value of \(a\) is
1 \(3250 \AA\)
2 \(6.5 \times 10^{-4} \mathrm{~mm}\)
3 1.24 micron
4 \(2.6 \times 10^{-4} \mathrm{~cm}\)
Explanation:
: Given that, \(\lambda=6200 \AA=6200 \times 10^{-10} \mathrm{~m}, \theta=\) \(30^{\circ}\) We know that, For first minima- \(\mathrm{a} \sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}=\frac{6200 \times 10^{-10}}{\sin 30^{\circ}}\) \(\mathrm{a}=1.24 \times 10^{-6} \mathrm{~m}\) \(\mathrm{a}=1.24 \mu\)
UPSEE - 2010
WAVE OPTICS
283212
In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda=\) wavelength of the light \()\) is \(I\). If \(I_0\) denotes the maximum intensity, then \(\frac{I}{I_0}\) is equal to
1 \(\frac{1}{2}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{3}{4}\)
5 \(\frac{1}{\sqrt{3}}\)
Explanation:
: Given, \(\Delta x=\frac{\lambda}{6}\) We know, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\) \(\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}\) \(\Delta \phi=\frac{\pi}{3}\) Now, \(\quad I^{\prime}=I_1+I_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi\) And, \(\quad I^{\prime}=I+I+2 I \cos \frac{\pi}{3}=3 I\) And, \(\quad \mathrm{I}_{\mathrm{o}}=\mathrm{I}+\mathrm{I}+2 \mathrm{I} \cos 0^{\circ}=4 \mathrm{I}\) From equation (i) and (ii), we get- \(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}=\frac{3}{4}\)
283205
Red light of wavelength \(625 \mathrm{~nm}\) is incident normally on an optical diffraction grating with \(2 \times 10^5\) lines \(/ \mathrm{m}\). Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?
1 15
2 17
3 20
4 18
Explanation:
: We know that, For principle maxima in grating spectra - \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=1,2,3\) is the order of principle maximum and \(\theta\) is the angle of diffraction. So, \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 10^5}=8\) Now, number of maxima \(=(2 n+1)\) \(=(2 \times 8+1)=17\)
UPSEE - 2014
WAVE OPTICS
283206
A double slit experiment is performed with light of wavelength \(500 \mathrm{~nm}\). A thin film of thickness \(2 \mu \mathrm{m}\) and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
1 remain unshifted
2 shift downward by nearly two fringes
3 shift upward by nearly two fringes
4 shift downward by its fringes
Explanation:
: Given that, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{t}=2 \mu \mathrm{m}=2 \times 10^{-6} \mathrm{~m}\) Refractive index \((\mu)=1.5\) \(\mathrm{t}=2 \mu \mathrm{m}\) Extra path difference \(=(\mu-1) t\) Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\beta\) is unchanged, path difference will be shifted. So, \(\quad(\mu-1) \mathrm{t}=\mathrm{n} \lambda\) \(\mathrm{n}=\frac{(\mu-1) \mathrm{t}}{\lambda}=\frac{(1.5-1) \times 2 \times 10^{-6}}{500 \times 10^{-9}}=2\)Light will be shifted upward by nearly two fringes.
UPSEE - 2014
WAVE OPTICS
283208
A slit of width a is illuminated by white light. For red light \((\lambda=6200 \AA)\), the first minima is obtained at a diffraction angle of \(30^{\circ}\). Then the value of \(a\) is
1 \(3250 \AA\)
2 \(6.5 \times 10^{-4} \mathrm{~mm}\)
3 1.24 micron
4 \(2.6 \times 10^{-4} \mathrm{~cm}\)
Explanation:
: Given that, \(\lambda=6200 \AA=6200 \times 10^{-10} \mathrm{~m}, \theta=\) \(30^{\circ}\) We know that, For first minima- \(\mathrm{a} \sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}=\frac{6200 \times 10^{-10}}{\sin 30^{\circ}}\) \(\mathrm{a}=1.24 \times 10^{-6} \mathrm{~m}\) \(\mathrm{a}=1.24 \mu\)
UPSEE - 2010
WAVE OPTICS
283212
In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda=\) wavelength of the light \()\) is \(I\). If \(I_0\) denotes the maximum intensity, then \(\frac{I}{I_0}\) is equal to
1 \(\frac{1}{2}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{3}{4}\)
5 \(\frac{1}{\sqrt{3}}\)
Explanation:
: Given, \(\Delta x=\frac{\lambda}{6}\) We know, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\) \(\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}\) \(\Delta \phi=\frac{\pi}{3}\) Now, \(\quad I^{\prime}=I_1+I_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi\) And, \(\quad I^{\prime}=I+I+2 I \cos \frac{\pi}{3}=3 I\) And, \(\quad \mathrm{I}_{\mathrm{o}}=\mathrm{I}+\mathrm{I}+2 \mathrm{I} \cos 0^{\circ}=4 \mathrm{I}\) From equation (i) and (ii), we get- \(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}=\frac{3}{4}\)
283205
Red light of wavelength \(625 \mathrm{~nm}\) is incident normally on an optical diffraction grating with \(2 \times 10^5\) lines \(/ \mathrm{m}\). Including central principal maxima, how many maxima may be observed on a screen which is far from the grating?
1 15
2 17
3 20
4 18
Explanation:
: We know that, For principle maxima in grating spectra - \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=1,2,3\) is the order of principle maximum and \(\theta\) is the angle of diffraction. So, \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 10^5}=8\) Now, number of maxima \(=(2 n+1)\) \(=(2 \times 8+1)=17\)
UPSEE - 2014
WAVE OPTICS
283206
A double slit experiment is performed with light of wavelength \(500 \mathrm{~nm}\). A thin film of thickness \(2 \mu \mathrm{m}\) and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
1 remain unshifted
2 shift downward by nearly two fringes
3 shift upward by nearly two fringes
4 shift downward by its fringes
Explanation:
: Given that, \(\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\) \(\mathrm{t}=2 \mu \mathrm{m}=2 \times 10^{-6} \mathrm{~m}\) Refractive index \((\mu)=1.5\) \(\mathrm{t}=2 \mu \mathrm{m}\) Extra path difference \(=(\mu-1) t\) Fringe width \((\beta)=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) \(\beta\) is unchanged, path difference will be shifted. So, \(\quad(\mu-1) \mathrm{t}=\mathrm{n} \lambda\) \(\mathrm{n}=\frac{(\mu-1) \mathrm{t}}{\lambda}=\frac{(1.5-1) \times 2 \times 10^{-6}}{500 \times 10^{-9}}=2\)Light will be shifted upward by nearly two fringes.
UPSEE - 2014
WAVE OPTICS
283208
A slit of width a is illuminated by white light. For red light \((\lambda=6200 \AA)\), the first minima is obtained at a diffraction angle of \(30^{\circ}\). Then the value of \(a\) is
1 \(3250 \AA\)
2 \(6.5 \times 10^{-4} \mathrm{~mm}\)
3 1.24 micron
4 \(2.6 \times 10^{-4} \mathrm{~cm}\)
Explanation:
: Given that, \(\lambda=6200 \AA=6200 \times 10^{-10} \mathrm{~m}, \theta=\) \(30^{\circ}\) We know that, For first minima- \(\mathrm{a} \sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}=\frac{6200 \times 10^{-10}}{\sin 30^{\circ}}\) \(\mathrm{a}=1.24 \times 10^{-6} \mathrm{~m}\) \(\mathrm{a}=1.24 \mu\)
UPSEE - 2010
WAVE OPTICS
283212
In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda=\) wavelength of the light \()\) is \(I\). If \(I_0\) denotes the maximum intensity, then \(\frac{I}{I_0}\) is equal to
1 \(\frac{1}{2}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{3}{4}\)
5 \(\frac{1}{\sqrt{3}}\)
Explanation:
: Given, \(\Delta x=\frac{\lambda}{6}\) We know, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}\) \(\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}\) \(\Delta \phi=\frac{\pi}{3}\) Now, \(\quad I^{\prime}=I_1+I_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi\) And, \(\quad I^{\prime}=I+I+2 I \cos \frac{\pi}{3}=3 I\) And, \(\quad \mathrm{I}_{\mathrm{o}}=\mathrm{I}+\mathrm{I}+2 \mathrm{I} \cos 0^{\circ}=4 \mathrm{I}\) From equation (i) and (ii), we get- \(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{o}}}=\frac{3}{4}\)
