283491
In the original arrangement, the slits are illuminated by white light and a thin glass plate is introduced in the path of the slit \(A\). Then
1 central white fringe is shifted towards slit B and flanked by coloured fringes. violet to red in that order
2 central white fringe is shifted towards slit B and is flanked by coloured fringes, red to violet in that order
3 central white fringe is shifted towards slit A and is flanked by coloured fringes, red to violet in that order
4 central white fringe is shifted towards slit A and is flanked by coloured fringes, violet to red in that order.
Explanation:
: Since thin glass plate is introduced in the path of the slit A then central white fringe is shifted towards slit A. From, \(\mathrm{y}=\frac{\mathrm{D} \lambda}{\mathrm{d}}\) We know wavelength of violet light is minimum and red light is maximum Hence, coloured fringes will be in the order of violet to red.
COMEDK 2011
WAVE OPTICS
283492
Diffraction effects are more easily detected in the case of sound waves than light wave because
1 sound waves are longitudinal
2 sound waves have smaller wavelength
3 sound waves have larger wavelength
4 sound waves are transverse
Explanation:
: Wavelength of sound waves is quite large compared to dimensions of the obstacle. Therefore, sound wave can bend around obstacle, the wavelength of light wave is much smalls than dimensions of the obstacle. Therefore, light wave do not show diffraction that easily.
COMEDK 2012
WAVE OPTICS
283493
If \(\theta\) is polarizing angle for a medium in which the speed of light is \(v\) then, according to Brewster's Law
1 \(\theta=\sin ^{-1}(c / v)\)
2 \(\theta=\tan ^{-1}(\mathrm{c} / \mathrm{v})\)
3 \(\theta=\cos ^{-1}(\mathrm{c} / \mathrm{v})\)
4 \(\theta=\sin ^{-1}(\mathrm{v} / \mathrm{c})\)
Explanation:
: We know that, \(\tan \theta=\mu \quad(\because \mu=\) refractive index \()\) \(\mu=\frac{\mathrm{c}}{\mathrm{v}}\) Where, \(\mathrm{c}=\) speed of light in vacuum \(\tan \theta =\frac{\mathrm{c}}{\mathrm{v}}\) \(\theta =\tan ^{-1}\left(\frac{\mathrm{c}}{\mathrm{v}}\right)\)\(\mathrm{v}=\) speed of light in medium,
COMEDK 2012
WAVE OPTICS
283495
The correct relation between \(S, \theta\) and \(C\) for an optically active solution is
1 \(\mathrm{S}=\theta \mathrm{LC}\)
2 \(\theta=\) SLC
3 \(\mathrm{L}=\theta \mathrm{SC}\)
4 \(\mathrm{C}=\theta \mathrm{LS}\)
Explanation:
: For an optically active solution, the angle by which the plane polarized light is rotated is given by Where, \(\theta=\mathrm{SLC}\) \(\theta\) - rotated angle \(\mathrm{S}\) - specific rotation \(\mathrm{L}\) - Path length of polarimeter \(\mathrm{C}-\) concentration of solution
283491
In the original arrangement, the slits are illuminated by white light and a thin glass plate is introduced in the path of the slit \(A\). Then
1 central white fringe is shifted towards slit B and flanked by coloured fringes. violet to red in that order
2 central white fringe is shifted towards slit B and is flanked by coloured fringes, red to violet in that order
3 central white fringe is shifted towards slit A and is flanked by coloured fringes, red to violet in that order
4 central white fringe is shifted towards slit A and is flanked by coloured fringes, violet to red in that order.
Explanation:
: Since thin glass plate is introduced in the path of the slit A then central white fringe is shifted towards slit A. From, \(\mathrm{y}=\frac{\mathrm{D} \lambda}{\mathrm{d}}\) We know wavelength of violet light is minimum and red light is maximum Hence, coloured fringes will be in the order of violet to red.
COMEDK 2011
WAVE OPTICS
283492
Diffraction effects are more easily detected in the case of sound waves than light wave because
1 sound waves are longitudinal
2 sound waves have smaller wavelength
3 sound waves have larger wavelength
4 sound waves are transverse
Explanation:
: Wavelength of sound waves is quite large compared to dimensions of the obstacle. Therefore, sound wave can bend around obstacle, the wavelength of light wave is much smalls than dimensions of the obstacle. Therefore, light wave do not show diffraction that easily.
COMEDK 2012
WAVE OPTICS
283493
If \(\theta\) is polarizing angle for a medium in which the speed of light is \(v\) then, according to Brewster's Law
1 \(\theta=\sin ^{-1}(c / v)\)
2 \(\theta=\tan ^{-1}(\mathrm{c} / \mathrm{v})\)
3 \(\theta=\cos ^{-1}(\mathrm{c} / \mathrm{v})\)
4 \(\theta=\sin ^{-1}(\mathrm{v} / \mathrm{c})\)
Explanation:
: We know that, \(\tan \theta=\mu \quad(\because \mu=\) refractive index \()\) \(\mu=\frac{\mathrm{c}}{\mathrm{v}}\) Where, \(\mathrm{c}=\) speed of light in vacuum \(\tan \theta =\frac{\mathrm{c}}{\mathrm{v}}\) \(\theta =\tan ^{-1}\left(\frac{\mathrm{c}}{\mathrm{v}}\right)\)\(\mathrm{v}=\) speed of light in medium,
COMEDK 2012
WAVE OPTICS
283495
The correct relation between \(S, \theta\) and \(C\) for an optically active solution is
1 \(\mathrm{S}=\theta \mathrm{LC}\)
2 \(\theta=\) SLC
3 \(\mathrm{L}=\theta \mathrm{SC}\)
4 \(\mathrm{C}=\theta \mathrm{LS}\)
Explanation:
: For an optically active solution, the angle by which the plane polarized light is rotated is given by Where, \(\theta=\mathrm{SLC}\) \(\theta\) - rotated angle \(\mathrm{S}\) - specific rotation \(\mathrm{L}\) - Path length of polarimeter \(\mathrm{C}-\) concentration of solution
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WAVE OPTICS
283491
In the original arrangement, the slits are illuminated by white light and a thin glass plate is introduced in the path of the slit \(A\). Then
1 central white fringe is shifted towards slit B and flanked by coloured fringes. violet to red in that order
2 central white fringe is shifted towards slit B and is flanked by coloured fringes, red to violet in that order
3 central white fringe is shifted towards slit A and is flanked by coloured fringes, red to violet in that order
4 central white fringe is shifted towards slit A and is flanked by coloured fringes, violet to red in that order.
Explanation:
: Since thin glass plate is introduced in the path of the slit A then central white fringe is shifted towards slit A. From, \(\mathrm{y}=\frac{\mathrm{D} \lambda}{\mathrm{d}}\) We know wavelength of violet light is minimum and red light is maximum Hence, coloured fringes will be in the order of violet to red.
COMEDK 2011
WAVE OPTICS
283492
Diffraction effects are more easily detected in the case of sound waves than light wave because
1 sound waves are longitudinal
2 sound waves have smaller wavelength
3 sound waves have larger wavelength
4 sound waves are transverse
Explanation:
: Wavelength of sound waves is quite large compared to dimensions of the obstacle. Therefore, sound wave can bend around obstacle, the wavelength of light wave is much smalls than dimensions of the obstacle. Therefore, light wave do not show diffraction that easily.
COMEDK 2012
WAVE OPTICS
283493
If \(\theta\) is polarizing angle for a medium in which the speed of light is \(v\) then, according to Brewster's Law
1 \(\theta=\sin ^{-1}(c / v)\)
2 \(\theta=\tan ^{-1}(\mathrm{c} / \mathrm{v})\)
3 \(\theta=\cos ^{-1}(\mathrm{c} / \mathrm{v})\)
4 \(\theta=\sin ^{-1}(\mathrm{v} / \mathrm{c})\)
Explanation:
: We know that, \(\tan \theta=\mu \quad(\because \mu=\) refractive index \()\) \(\mu=\frac{\mathrm{c}}{\mathrm{v}}\) Where, \(\mathrm{c}=\) speed of light in vacuum \(\tan \theta =\frac{\mathrm{c}}{\mathrm{v}}\) \(\theta =\tan ^{-1}\left(\frac{\mathrm{c}}{\mathrm{v}}\right)\)\(\mathrm{v}=\) speed of light in medium,
COMEDK 2012
WAVE OPTICS
283495
The correct relation between \(S, \theta\) and \(C\) for an optically active solution is
1 \(\mathrm{S}=\theta \mathrm{LC}\)
2 \(\theta=\) SLC
3 \(\mathrm{L}=\theta \mathrm{SC}\)
4 \(\mathrm{C}=\theta \mathrm{LS}\)
Explanation:
: For an optically active solution, the angle by which the plane polarized light is rotated is given by Where, \(\theta=\mathrm{SLC}\) \(\theta\) - rotated angle \(\mathrm{S}\) - specific rotation \(\mathrm{L}\) - Path length of polarimeter \(\mathrm{C}-\) concentration of solution
283491
In the original arrangement, the slits are illuminated by white light and a thin glass plate is introduced in the path of the slit \(A\). Then
1 central white fringe is shifted towards slit B and flanked by coloured fringes. violet to red in that order
2 central white fringe is shifted towards slit B and is flanked by coloured fringes, red to violet in that order
3 central white fringe is shifted towards slit A and is flanked by coloured fringes, red to violet in that order
4 central white fringe is shifted towards slit A and is flanked by coloured fringes, violet to red in that order.
Explanation:
: Since thin glass plate is introduced in the path of the slit A then central white fringe is shifted towards slit A. From, \(\mathrm{y}=\frac{\mathrm{D} \lambda}{\mathrm{d}}\) We know wavelength of violet light is minimum and red light is maximum Hence, coloured fringes will be in the order of violet to red.
COMEDK 2011
WAVE OPTICS
283492
Diffraction effects are more easily detected in the case of sound waves than light wave because
1 sound waves are longitudinal
2 sound waves have smaller wavelength
3 sound waves have larger wavelength
4 sound waves are transverse
Explanation:
: Wavelength of sound waves is quite large compared to dimensions of the obstacle. Therefore, sound wave can bend around obstacle, the wavelength of light wave is much smalls than dimensions of the obstacle. Therefore, light wave do not show diffraction that easily.
COMEDK 2012
WAVE OPTICS
283493
If \(\theta\) is polarizing angle for a medium in which the speed of light is \(v\) then, according to Brewster's Law
1 \(\theta=\sin ^{-1}(c / v)\)
2 \(\theta=\tan ^{-1}(\mathrm{c} / \mathrm{v})\)
3 \(\theta=\cos ^{-1}(\mathrm{c} / \mathrm{v})\)
4 \(\theta=\sin ^{-1}(\mathrm{v} / \mathrm{c})\)
Explanation:
: We know that, \(\tan \theta=\mu \quad(\because \mu=\) refractive index \()\) \(\mu=\frac{\mathrm{c}}{\mathrm{v}}\) Where, \(\mathrm{c}=\) speed of light in vacuum \(\tan \theta =\frac{\mathrm{c}}{\mathrm{v}}\) \(\theta =\tan ^{-1}\left(\frac{\mathrm{c}}{\mathrm{v}}\right)\)\(\mathrm{v}=\) speed of light in medium,
COMEDK 2012
WAVE OPTICS
283495
The correct relation between \(S, \theta\) and \(C\) for an optically active solution is
1 \(\mathrm{S}=\theta \mathrm{LC}\)
2 \(\theta=\) SLC
3 \(\mathrm{L}=\theta \mathrm{SC}\)
4 \(\mathrm{C}=\theta \mathrm{LS}\)
Explanation:
: For an optically active solution, the angle by which the plane polarized light is rotated is given by Where, \(\theta=\mathrm{SLC}\) \(\theta\) - rotated angle \(\mathrm{S}\) - specific rotation \(\mathrm{L}\) - Path length of polarimeter \(\mathrm{C}-\) concentration of solution