283613
The angular resolution of a \(10 \mathrm{~cm}\) diameter telescope at a wavelength of \(5000 \AA\) is of the order of
1 \(10^6 \mathrm{rad}\)
2 \(10^{-2} \mathrm{rad}\)
3 \(10^{-4} \mathrm{rad}\)
4 \(10^{-6} \mathrm{rad}\)
Explanation:
: Given that, \(\mathrm{d}=10 \mathrm{~cm}=0.10, \lambda=5000 \AA= 5000 \times 10^{-10} \mathrm{~m}\) \(= 5 \times 10^{-7} \mathrm{~m}\) We know that, Angular resolution \(\theta=\frac{1.22 \lambda}{\mathrm{d}}\) Substituting the value of \(\lambda\) and \(d\) in equation (i), we get \(\theta=\frac{1.22 \times 5 \times 10^{-7}}{0.10}\) \(\frac{122 \times 5 \times 10^{-7}}{10}\) \(61 \times 10^{-7}\) \(6.1 \times 10^{-6} \text { radian }\)\(\therefore\) The order will be of \(10^{-6} \mathrm{rad}\).
AIPMT - 2005
WAVE OPTICS
283618
Unpolarized light falls first on polarizer (P) and then on analyser (A). If the intensity of the transmitted light from the analyser is \(\frac{1}{8}\) th of the incident unpolarised light. What will be the angle between optic axes of \(P \& A\) ?
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(0^0\)
4 \(60^{\circ}\)
Explanation:
: Given, \(\mathrm{I}=\frac{\mathrm{I}_{\mathrm{o}}}{2}\) Intensity of transmitted light, \(\mathrm{I}=\frac{1}{8} \quad \mathrm{I}_0\) Where, \(\mathrm{I}_0=\) incident unpolarised light Now intensity of transmitted light \(I=\frac{I_0}{2} \cos ^2 \theta\) \(\frac{1}{8} I_0=\frac{I_0}{2} \cos ^2 \theta\) \(\cos ^2 \theta=\frac{1}{4}\) \(\cos \theta=\frac{1}{2}=\cos 60^{\circ}\) \(\theta=60^{\circ}\)Hence, the angle between option axes of \(\mathrm{P}\) and \(\mathrm{A}\) is \(60^{\circ}\).
GUJCET 2014
WAVE OPTICS
283621
When the angle of incidence is \(60^{\circ}\) on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
1 \(\sqrt{2} \times 10^8 \mathrm{~ms}^{-1}\)
2 \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
3 \(2 \times 10^8 \mathrm{~ms}^{-1}\)
4 \(3 \times 10^8 \mathrm{~ms}^{-1}\)
Explanation:
: Given, \(\mathrm{i}=60^{\circ}\) \(\mathrm{i}=\mathrm{i}_{\mathrm{p}}\) (because it is completely polarized). According to Brewster's law - \(\mu=\tan i_p=\frac{\text { speed of light in air }}{\text { speed of light in medium }}\) \(\tan 60^{\circ}=\frac{3 \times 10^8}{\mathrm{v}}\) \(\mathrm{v}=\frac{3 \times 10^8}{\sqrt{3}}\) \(\mathrm{v}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
JCECE-2014
WAVE OPTICS
283622
For a transparent medium the polarising angle is \(60^{\circ}\). What is the angle of refraction?
283613
The angular resolution of a \(10 \mathrm{~cm}\) diameter telescope at a wavelength of \(5000 \AA\) is of the order of
1 \(10^6 \mathrm{rad}\)
2 \(10^{-2} \mathrm{rad}\)
3 \(10^{-4} \mathrm{rad}\)
4 \(10^{-6} \mathrm{rad}\)
Explanation:
: Given that, \(\mathrm{d}=10 \mathrm{~cm}=0.10, \lambda=5000 \AA= 5000 \times 10^{-10} \mathrm{~m}\) \(= 5 \times 10^{-7} \mathrm{~m}\) We know that, Angular resolution \(\theta=\frac{1.22 \lambda}{\mathrm{d}}\) Substituting the value of \(\lambda\) and \(d\) in equation (i), we get \(\theta=\frac{1.22 \times 5 \times 10^{-7}}{0.10}\) \(\frac{122 \times 5 \times 10^{-7}}{10}\) \(61 \times 10^{-7}\) \(6.1 \times 10^{-6} \text { radian }\)\(\therefore\) The order will be of \(10^{-6} \mathrm{rad}\).
AIPMT - 2005
WAVE OPTICS
283618
Unpolarized light falls first on polarizer (P) and then on analyser (A). If the intensity of the transmitted light from the analyser is \(\frac{1}{8}\) th of the incident unpolarised light. What will be the angle between optic axes of \(P \& A\) ?
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(0^0\)
4 \(60^{\circ}\)
Explanation:
: Given, \(\mathrm{I}=\frac{\mathrm{I}_{\mathrm{o}}}{2}\) Intensity of transmitted light, \(\mathrm{I}=\frac{1}{8} \quad \mathrm{I}_0\) Where, \(\mathrm{I}_0=\) incident unpolarised light Now intensity of transmitted light \(I=\frac{I_0}{2} \cos ^2 \theta\) \(\frac{1}{8} I_0=\frac{I_0}{2} \cos ^2 \theta\) \(\cos ^2 \theta=\frac{1}{4}\) \(\cos \theta=\frac{1}{2}=\cos 60^{\circ}\) \(\theta=60^{\circ}\)Hence, the angle between option axes of \(\mathrm{P}\) and \(\mathrm{A}\) is \(60^{\circ}\).
GUJCET 2014
WAVE OPTICS
283621
When the angle of incidence is \(60^{\circ}\) on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
1 \(\sqrt{2} \times 10^8 \mathrm{~ms}^{-1}\)
2 \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
3 \(2 \times 10^8 \mathrm{~ms}^{-1}\)
4 \(3 \times 10^8 \mathrm{~ms}^{-1}\)
Explanation:
: Given, \(\mathrm{i}=60^{\circ}\) \(\mathrm{i}=\mathrm{i}_{\mathrm{p}}\) (because it is completely polarized). According to Brewster's law - \(\mu=\tan i_p=\frac{\text { speed of light in air }}{\text { speed of light in medium }}\) \(\tan 60^{\circ}=\frac{3 \times 10^8}{\mathrm{v}}\) \(\mathrm{v}=\frac{3 \times 10^8}{\sqrt{3}}\) \(\mathrm{v}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
JCECE-2014
WAVE OPTICS
283622
For a transparent medium the polarising angle is \(60^{\circ}\). What is the angle of refraction?
283613
The angular resolution of a \(10 \mathrm{~cm}\) diameter telescope at a wavelength of \(5000 \AA\) is of the order of
1 \(10^6 \mathrm{rad}\)
2 \(10^{-2} \mathrm{rad}\)
3 \(10^{-4} \mathrm{rad}\)
4 \(10^{-6} \mathrm{rad}\)
Explanation:
: Given that, \(\mathrm{d}=10 \mathrm{~cm}=0.10, \lambda=5000 \AA= 5000 \times 10^{-10} \mathrm{~m}\) \(= 5 \times 10^{-7} \mathrm{~m}\) We know that, Angular resolution \(\theta=\frac{1.22 \lambda}{\mathrm{d}}\) Substituting the value of \(\lambda\) and \(d\) in equation (i), we get \(\theta=\frac{1.22 \times 5 \times 10^{-7}}{0.10}\) \(\frac{122 \times 5 \times 10^{-7}}{10}\) \(61 \times 10^{-7}\) \(6.1 \times 10^{-6} \text { radian }\)\(\therefore\) The order will be of \(10^{-6} \mathrm{rad}\).
AIPMT - 2005
WAVE OPTICS
283618
Unpolarized light falls first on polarizer (P) and then on analyser (A). If the intensity of the transmitted light from the analyser is \(\frac{1}{8}\) th of the incident unpolarised light. What will be the angle between optic axes of \(P \& A\) ?
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(0^0\)
4 \(60^{\circ}\)
Explanation:
: Given, \(\mathrm{I}=\frac{\mathrm{I}_{\mathrm{o}}}{2}\) Intensity of transmitted light, \(\mathrm{I}=\frac{1}{8} \quad \mathrm{I}_0\) Where, \(\mathrm{I}_0=\) incident unpolarised light Now intensity of transmitted light \(I=\frac{I_0}{2} \cos ^2 \theta\) \(\frac{1}{8} I_0=\frac{I_0}{2} \cos ^2 \theta\) \(\cos ^2 \theta=\frac{1}{4}\) \(\cos \theta=\frac{1}{2}=\cos 60^{\circ}\) \(\theta=60^{\circ}\)Hence, the angle between option axes of \(\mathrm{P}\) and \(\mathrm{A}\) is \(60^{\circ}\).
GUJCET 2014
WAVE OPTICS
283621
When the angle of incidence is \(60^{\circ}\) on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
1 \(\sqrt{2} \times 10^8 \mathrm{~ms}^{-1}\)
2 \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
3 \(2 \times 10^8 \mathrm{~ms}^{-1}\)
4 \(3 \times 10^8 \mathrm{~ms}^{-1}\)
Explanation:
: Given, \(\mathrm{i}=60^{\circ}\) \(\mathrm{i}=\mathrm{i}_{\mathrm{p}}\) (because it is completely polarized). According to Brewster's law - \(\mu=\tan i_p=\frac{\text { speed of light in air }}{\text { speed of light in medium }}\) \(\tan 60^{\circ}=\frac{3 \times 10^8}{\mathrm{v}}\) \(\mathrm{v}=\frac{3 \times 10^8}{\sqrt{3}}\) \(\mathrm{v}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
JCECE-2014
WAVE OPTICS
283622
For a transparent medium the polarising angle is \(60^{\circ}\). What is the angle of refraction?
283613
The angular resolution of a \(10 \mathrm{~cm}\) diameter telescope at a wavelength of \(5000 \AA\) is of the order of
1 \(10^6 \mathrm{rad}\)
2 \(10^{-2} \mathrm{rad}\)
3 \(10^{-4} \mathrm{rad}\)
4 \(10^{-6} \mathrm{rad}\)
Explanation:
: Given that, \(\mathrm{d}=10 \mathrm{~cm}=0.10, \lambda=5000 \AA= 5000 \times 10^{-10} \mathrm{~m}\) \(= 5 \times 10^{-7} \mathrm{~m}\) We know that, Angular resolution \(\theta=\frac{1.22 \lambda}{\mathrm{d}}\) Substituting the value of \(\lambda\) and \(d\) in equation (i), we get \(\theta=\frac{1.22 \times 5 \times 10^{-7}}{0.10}\) \(\frac{122 \times 5 \times 10^{-7}}{10}\) \(61 \times 10^{-7}\) \(6.1 \times 10^{-6} \text { radian }\)\(\therefore\) The order will be of \(10^{-6} \mathrm{rad}\).
AIPMT - 2005
WAVE OPTICS
283618
Unpolarized light falls first on polarizer (P) and then on analyser (A). If the intensity of the transmitted light from the analyser is \(\frac{1}{8}\) th of the incident unpolarised light. What will be the angle between optic axes of \(P \& A\) ?
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(0^0\)
4 \(60^{\circ}\)
Explanation:
: Given, \(\mathrm{I}=\frac{\mathrm{I}_{\mathrm{o}}}{2}\) Intensity of transmitted light, \(\mathrm{I}=\frac{1}{8} \quad \mathrm{I}_0\) Where, \(\mathrm{I}_0=\) incident unpolarised light Now intensity of transmitted light \(I=\frac{I_0}{2} \cos ^2 \theta\) \(\frac{1}{8} I_0=\frac{I_0}{2} \cos ^2 \theta\) \(\cos ^2 \theta=\frac{1}{4}\) \(\cos \theta=\frac{1}{2}=\cos 60^{\circ}\) \(\theta=60^{\circ}\)Hence, the angle between option axes of \(\mathrm{P}\) and \(\mathrm{A}\) is \(60^{\circ}\).
GUJCET 2014
WAVE OPTICS
283621
When the angle of incidence is \(60^{\circ}\) on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
1 \(\sqrt{2} \times 10^8 \mathrm{~ms}^{-1}\)
2 \(\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}\)
3 \(2 \times 10^8 \mathrm{~ms}^{-1}\)
4 \(3 \times 10^8 \mathrm{~ms}^{-1}\)
Explanation:
: Given, \(\mathrm{i}=60^{\circ}\) \(\mathrm{i}=\mathrm{i}_{\mathrm{p}}\) (because it is completely polarized). According to Brewster's law - \(\mu=\tan i_p=\frac{\text { speed of light in air }}{\text { speed of light in medium }}\) \(\tan 60^{\circ}=\frac{3 \times 10^8}{\mathrm{v}}\) \(\mathrm{v}=\frac{3 \times 10^8}{\sqrt{3}}\) \(\mathrm{v}=\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}\)
JCECE-2014
WAVE OPTICS
283622
For a transparent medium the polarising angle is \(60^{\circ}\). What is the angle of refraction?
