283516
From Brewster's law, except for polished metallic surface, the polarising angle
1 depends on wavelength and is different for different colours
2 independent of wavelength and is different for different colours
3 independent of wavelength and is same for different colours
4 depends on wavelength and is same for different colours
Explanation:
: As we know, according to Brewster's law \(\tan \mathrm{i}=\mu\) Where, \(\mathrm{i}=\tan ^{-1} \mu\) \(\mathrm{i}=\) Polarisation angle \(\mu=\) Refractive index So, we can say that the polarising angle depends on refractive index. And the refractive index is different for different wavelength and therefore, is different for different colour.
MHT-CET 2016
WAVE OPTICS
283523
A slit of width ' \(a\) ' is illuminated with a monochromatic light of wavelength \(\lambda\) from a distant source and the diffraction pattern is observed on a screen placed at a distance ' \(D\) ' from the slit. To increase the width of the central maximum one should
1 decrease D
2 decrease a
3 decrease \(\lambda\)
4 the width cannot be changed
Explanation:
: By the theory of diffraction at a single slit, the width of the central maximum is given by \(\omega=\frac{2 \mathrm{D} \lambda}{\mathrm{a}}\) \(\omega \propto \mathrm{D}\) \(\omega \propto \lambda\) \(\omega \propto \frac{1}{\mathrm{a}}\) Therefore, to increases the width of the central maximum 'a' should be decreased.
VITEEE-2008
WAVE OPTICS
283529
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 8
4 16
Explanation:
: For principal maxima in grating spectra \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=(1,2,3\), is the order or principal maxima and \(\theta\) is the angle of diffraction. \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 2 \times 10^5}=8\) Total number of maximum \(=2 \times\) number of minimum +1 \(=2 \times 8+1\) \(=16+1=17\)
Karnataka CET-2010
WAVE OPTICS
283530
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400 \mathrm{~nm}\), the first minimum is formed at an angle of \(30^{\circ}\). The direction \(\theta\) of the first secondary maximum is given by :
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
: For first diffraction minimum, a \(\sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}\) For first secondary maximum, \(a \sin \theta^{\prime}=\frac{3 \lambda}{2}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{1}{a}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{\sin \theta}{\lambda}\) \(\sin \theta^{\prime}=\frac{3}{2} \times \sin 30^{\circ}=\frac{3}{4}\) \(\theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)
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WAVE OPTICS
283516
From Brewster's law, except for polished metallic surface, the polarising angle
1 depends on wavelength and is different for different colours
2 independent of wavelength and is different for different colours
3 independent of wavelength and is same for different colours
4 depends on wavelength and is same for different colours
Explanation:
: As we know, according to Brewster's law \(\tan \mathrm{i}=\mu\) Where, \(\mathrm{i}=\tan ^{-1} \mu\) \(\mathrm{i}=\) Polarisation angle \(\mu=\) Refractive index So, we can say that the polarising angle depends on refractive index. And the refractive index is different for different wavelength and therefore, is different for different colour.
MHT-CET 2016
WAVE OPTICS
283523
A slit of width ' \(a\) ' is illuminated with a monochromatic light of wavelength \(\lambda\) from a distant source and the diffraction pattern is observed on a screen placed at a distance ' \(D\) ' from the slit. To increase the width of the central maximum one should
1 decrease D
2 decrease a
3 decrease \(\lambda\)
4 the width cannot be changed
Explanation:
: By the theory of diffraction at a single slit, the width of the central maximum is given by \(\omega=\frac{2 \mathrm{D} \lambda}{\mathrm{a}}\) \(\omega \propto \mathrm{D}\) \(\omega \propto \lambda\) \(\omega \propto \frac{1}{\mathrm{a}}\) Therefore, to increases the width of the central maximum 'a' should be decreased.
VITEEE-2008
WAVE OPTICS
283529
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 8
4 16
Explanation:
: For principal maxima in grating spectra \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=(1,2,3\), is the order or principal maxima and \(\theta\) is the angle of diffraction. \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 2 \times 10^5}=8\) Total number of maximum \(=2 \times\) number of minimum +1 \(=2 \times 8+1\) \(=16+1=17\)
Karnataka CET-2010
WAVE OPTICS
283530
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400 \mathrm{~nm}\), the first minimum is formed at an angle of \(30^{\circ}\). The direction \(\theta\) of the first secondary maximum is given by :
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
: For first diffraction minimum, a \(\sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}\) For first secondary maximum, \(a \sin \theta^{\prime}=\frac{3 \lambda}{2}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{1}{a}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{\sin \theta}{\lambda}\) \(\sin \theta^{\prime}=\frac{3}{2} \times \sin 30^{\circ}=\frac{3}{4}\) \(\theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)
283516
From Brewster's law, except for polished metallic surface, the polarising angle
1 depends on wavelength and is different for different colours
2 independent of wavelength and is different for different colours
3 independent of wavelength and is same for different colours
4 depends on wavelength and is same for different colours
Explanation:
: As we know, according to Brewster's law \(\tan \mathrm{i}=\mu\) Where, \(\mathrm{i}=\tan ^{-1} \mu\) \(\mathrm{i}=\) Polarisation angle \(\mu=\) Refractive index So, we can say that the polarising angle depends on refractive index. And the refractive index is different for different wavelength and therefore, is different for different colour.
MHT-CET 2016
WAVE OPTICS
283523
A slit of width ' \(a\) ' is illuminated with a monochromatic light of wavelength \(\lambda\) from a distant source and the diffraction pattern is observed on a screen placed at a distance ' \(D\) ' from the slit. To increase the width of the central maximum one should
1 decrease D
2 decrease a
3 decrease \(\lambda\)
4 the width cannot be changed
Explanation:
: By the theory of diffraction at a single slit, the width of the central maximum is given by \(\omega=\frac{2 \mathrm{D} \lambda}{\mathrm{a}}\) \(\omega \propto \mathrm{D}\) \(\omega \propto \lambda\) \(\omega \propto \frac{1}{\mathrm{a}}\) Therefore, to increases the width of the central maximum 'a' should be decreased.
VITEEE-2008
WAVE OPTICS
283529
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 8
4 16
Explanation:
: For principal maxima in grating spectra \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=(1,2,3\), is the order or principal maxima and \(\theta\) is the angle of diffraction. \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 2 \times 10^5}=8\) Total number of maximum \(=2 \times\) number of minimum +1 \(=2 \times 8+1\) \(=16+1=17\)
Karnataka CET-2010
WAVE OPTICS
283530
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400 \mathrm{~nm}\), the first minimum is formed at an angle of \(30^{\circ}\). The direction \(\theta\) of the first secondary maximum is given by :
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
: For first diffraction minimum, a \(\sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}\) For first secondary maximum, \(a \sin \theta^{\prime}=\frac{3 \lambda}{2}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{1}{a}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{\sin \theta}{\lambda}\) \(\sin \theta^{\prime}=\frac{3}{2} \times \sin 30^{\circ}=\frac{3}{4}\) \(\theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)
283516
From Brewster's law, except for polished metallic surface, the polarising angle
1 depends on wavelength and is different for different colours
2 independent of wavelength and is different for different colours
3 independent of wavelength and is same for different colours
4 depends on wavelength and is same for different colours
Explanation:
: As we know, according to Brewster's law \(\tan \mathrm{i}=\mu\) Where, \(\mathrm{i}=\tan ^{-1} \mu\) \(\mathrm{i}=\) Polarisation angle \(\mu=\) Refractive index So, we can say that the polarising angle depends on refractive index. And the refractive index is different for different wavelength and therefore, is different for different colour.
MHT-CET 2016
WAVE OPTICS
283523
A slit of width ' \(a\) ' is illuminated with a monochromatic light of wavelength \(\lambda\) from a distant source and the diffraction pattern is observed on a screen placed at a distance ' \(D\) ' from the slit. To increase the width of the central maximum one should
1 decrease D
2 decrease a
3 decrease \(\lambda\)
4 the width cannot be changed
Explanation:
: By the theory of diffraction at a single slit, the width of the central maximum is given by \(\omega=\frac{2 \mathrm{D} \lambda}{\mathrm{a}}\) \(\omega \propto \mathrm{D}\) \(\omega \propto \lambda\) \(\omega \propto \frac{1}{\mathrm{a}}\) Therefore, to increases the width of the central maximum 'a' should be decreased.
VITEEE-2008
WAVE OPTICS
283529
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 8
4 16
Explanation:
: For principal maxima in grating spectra \(\frac{\sin \theta}{\mathrm{N}}=\mathrm{n} \lambda\) Where, \(\mathrm{n}=(1,2,3\), is the order or principal maxima and \(\theta\) is the angle of diffraction. \(\mathrm{n}=\frac{1}{\lambda \mathrm{N}}=\frac{1}{6.25 \times 10^{-7} \times 2 \times 2 \times 10^5}=8\) Total number of maximum \(=2 \times\) number of minimum +1 \(=2 \times 8+1\) \(=16+1=17\)
Karnataka CET-2010
WAVE OPTICS
283530
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400 \mathrm{~nm}\), the first minimum is formed at an angle of \(30^{\circ}\). The direction \(\theta\) of the first secondary maximum is given by :
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
: For first diffraction minimum, a \(\sin \theta=\lambda\) \(\mathrm{a}=\frac{\lambda}{\sin \theta}\) For first secondary maximum, \(a \sin \theta^{\prime}=\frac{3 \lambda}{2}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{1}{a}\) \(\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{\sin \theta}{\lambda}\) \(\sin \theta^{\prime}=\frac{3}{2} \times \sin 30^{\circ}=\frac{3}{4}\) \(\theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)