283594
A polarized light of intensity \(I_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emergent polarized light from second polarizer ?
1 \(\mathrm{I}=\mathrm{I}_0\)
2 \(\mathrm{I}=\mathrm{I}_0 / 6\)
3 \(\mathrm{I}=\mathrm{I}_0 / 5\)
4 \(\mathrm{I}_0 / 4\)
Explanation:
: Given that, \(\theta=60^{\circ}\) \(\mathrm{I}=? \quad(\mathrm{I}=\) Intensity passed through polariser \()\) According to the Malus's law - \(I=I_0 \cos ^2 \theta\) \(I=I_0 \cos ^2 60^{\circ}\) \(I=I_0 \times\left(\frac{1}{2}\right)^2\) \(I=\frac{I_0}{4}\)
Karnataka CET-2014
WAVE OPTICS
283595
Critical angle for certain medium is \(\sin ^{-1}(\mathbf{0 . 6})\). The polarizing angle of that medium is :
283597
An unpolarised beam of intensity \(I_0\) is incident on a pair of nicols making an angle of \(60^{\circ}\) with each other. The intensity of light emerging from the pair is :
1 \(\mathrm{I}_0\)
2 \(\mathrm{I}_0 / 2\)
3 \(\mathrm{I}_0 / 4\)
4 \(\mathrm{I}_0 / 8\)
Explanation:
: Given that, angle \((\theta)=60^{\circ}\) Intensity of light emerging from \(1^{\text {st }}\) prism - \(I_1=\frac{I_0}{2}\) Intensity of light emerging from \(2^{\text {nd }}\) prism - \(I_2=\frac{I_0}{2} \cos ^2 \theta\) \(I_2=\frac{I_0}{2} \cos ^2 60^{\circ}\) \(I_2=\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8}\)
Karnataka CET-2006
WAVE OPTICS
283600
The Brewster angle for the glass-air interface is \(54.74^{\circ}\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\)
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(25^{\circ}\)
4 \(54.74^{\circ}\)
Explanation:
: Given that, \(\mathrm{i}_{\mathrm{p}}=54.74^{\circ}\) According to the Brewster's law - \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\sqrt{2}=\mu\) \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\) According to the Snell's law - \(\mu=\frac{\sin i}{\sin r}\) \(\sqrt{2}=\frac{\sin i}{\sin r}\) Given that, \(\mathrm{i}=45^{\circ}\) \(\sqrt{2}=\frac{1}{\sqrt{2} \sin r}\) \(\sin r=\frac{1}{2}\) \(r=\sin ^{-1}\left(\frac{1}{2}\right)\) \(r=30^{\circ}\)
283594
A polarized light of intensity \(I_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emergent polarized light from second polarizer ?
1 \(\mathrm{I}=\mathrm{I}_0\)
2 \(\mathrm{I}=\mathrm{I}_0 / 6\)
3 \(\mathrm{I}=\mathrm{I}_0 / 5\)
4 \(\mathrm{I}_0 / 4\)
Explanation:
: Given that, \(\theta=60^{\circ}\) \(\mathrm{I}=? \quad(\mathrm{I}=\) Intensity passed through polariser \()\) According to the Malus's law - \(I=I_0 \cos ^2 \theta\) \(I=I_0 \cos ^2 60^{\circ}\) \(I=I_0 \times\left(\frac{1}{2}\right)^2\) \(I=\frac{I_0}{4}\)
Karnataka CET-2014
WAVE OPTICS
283595
Critical angle for certain medium is \(\sin ^{-1}(\mathbf{0 . 6})\). The polarizing angle of that medium is :
283597
An unpolarised beam of intensity \(I_0\) is incident on a pair of nicols making an angle of \(60^{\circ}\) with each other. The intensity of light emerging from the pair is :
1 \(\mathrm{I}_0\)
2 \(\mathrm{I}_0 / 2\)
3 \(\mathrm{I}_0 / 4\)
4 \(\mathrm{I}_0 / 8\)
Explanation:
: Given that, angle \((\theta)=60^{\circ}\) Intensity of light emerging from \(1^{\text {st }}\) prism - \(I_1=\frac{I_0}{2}\) Intensity of light emerging from \(2^{\text {nd }}\) prism - \(I_2=\frac{I_0}{2} \cos ^2 \theta\) \(I_2=\frac{I_0}{2} \cos ^2 60^{\circ}\) \(I_2=\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8}\)
Karnataka CET-2006
WAVE OPTICS
283600
The Brewster angle for the glass-air interface is \(54.74^{\circ}\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\)
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(25^{\circ}\)
4 \(54.74^{\circ}\)
Explanation:
: Given that, \(\mathrm{i}_{\mathrm{p}}=54.74^{\circ}\) According to the Brewster's law - \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\sqrt{2}=\mu\) \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\) According to the Snell's law - \(\mu=\frac{\sin i}{\sin r}\) \(\sqrt{2}=\frac{\sin i}{\sin r}\) Given that, \(\mathrm{i}=45^{\circ}\) \(\sqrt{2}=\frac{1}{\sqrt{2} \sin r}\) \(\sin r=\frac{1}{2}\) \(r=\sin ^{-1}\left(\frac{1}{2}\right)\) \(r=30^{\circ}\)
283594
A polarized light of intensity \(I_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emergent polarized light from second polarizer ?
1 \(\mathrm{I}=\mathrm{I}_0\)
2 \(\mathrm{I}=\mathrm{I}_0 / 6\)
3 \(\mathrm{I}=\mathrm{I}_0 / 5\)
4 \(\mathrm{I}_0 / 4\)
Explanation:
: Given that, \(\theta=60^{\circ}\) \(\mathrm{I}=? \quad(\mathrm{I}=\) Intensity passed through polariser \()\) According to the Malus's law - \(I=I_0 \cos ^2 \theta\) \(I=I_0 \cos ^2 60^{\circ}\) \(I=I_0 \times\left(\frac{1}{2}\right)^2\) \(I=\frac{I_0}{4}\)
Karnataka CET-2014
WAVE OPTICS
283595
Critical angle for certain medium is \(\sin ^{-1}(\mathbf{0 . 6})\). The polarizing angle of that medium is :
283597
An unpolarised beam of intensity \(I_0\) is incident on a pair of nicols making an angle of \(60^{\circ}\) with each other. The intensity of light emerging from the pair is :
1 \(\mathrm{I}_0\)
2 \(\mathrm{I}_0 / 2\)
3 \(\mathrm{I}_0 / 4\)
4 \(\mathrm{I}_0 / 8\)
Explanation:
: Given that, angle \((\theta)=60^{\circ}\) Intensity of light emerging from \(1^{\text {st }}\) prism - \(I_1=\frac{I_0}{2}\) Intensity of light emerging from \(2^{\text {nd }}\) prism - \(I_2=\frac{I_0}{2} \cos ^2 \theta\) \(I_2=\frac{I_0}{2} \cos ^2 60^{\circ}\) \(I_2=\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8}\)
Karnataka CET-2006
WAVE OPTICS
283600
The Brewster angle for the glass-air interface is \(54.74^{\circ}\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\)
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(25^{\circ}\)
4 \(54.74^{\circ}\)
Explanation:
: Given that, \(\mathrm{i}_{\mathrm{p}}=54.74^{\circ}\) According to the Brewster's law - \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\sqrt{2}=\mu\) \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\) According to the Snell's law - \(\mu=\frac{\sin i}{\sin r}\) \(\sqrt{2}=\frac{\sin i}{\sin r}\) Given that, \(\mathrm{i}=45^{\circ}\) \(\sqrt{2}=\frac{1}{\sqrt{2} \sin r}\) \(\sin r=\frac{1}{2}\) \(r=\sin ^{-1}\left(\frac{1}{2}\right)\) \(r=30^{\circ}\)
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WAVE OPTICS
283594
A polarized light of intensity \(I_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emergent polarized light from second polarizer ?
1 \(\mathrm{I}=\mathrm{I}_0\)
2 \(\mathrm{I}=\mathrm{I}_0 / 6\)
3 \(\mathrm{I}=\mathrm{I}_0 / 5\)
4 \(\mathrm{I}_0 / 4\)
Explanation:
: Given that, \(\theta=60^{\circ}\) \(\mathrm{I}=? \quad(\mathrm{I}=\) Intensity passed through polariser \()\) According to the Malus's law - \(I=I_0 \cos ^2 \theta\) \(I=I_0 \cos ^2 60^{\circ}\) \(I=I_0 \times\left(\frac{1}{2}\right)^2\) \(I=\frac{I_0}{4}\)
Karnataka CET-2014
WAVE OPTICS
283595
Critical angle for certain medium is \(\sin ^{-1}(\mathbf{0 . 6})\). The polarizing angle of that medium is :
283597
An unpolarised beam of intensity \(I_0\) is incident on a pair of nicols making an angle of \(60^{\circ}\) with each other. The intensity of light emerging from the pair is :
1 \(\mathrm{I}_0\)
2 \(\mathrm{I}_0 / 2\)
3 \(\mathrm{I}_0 / 4\)
4 \(\mathrm{I}_0 / 8\)
Explanation:
: Given that, angle \((\theta)=60^{\circ}\) Intensity of light emerging from \(1^{\text {st }}\) prism - \(I_1=\frac{I_0}{2}\) Intensity of light emerging from \(2^{\text {nd }}\) prism - \(I_2=\frac{I_0}{2} \cos ^2 \theta\) \(I_2=\frac{I_0}{2} \cos ^2 60^{\circ}\) \(I_2=\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8}\)
Karnataka CET-2006
WAVE OPTICS
283600
The Brewster angle for the glass-air interface is \(54.74^{\circ}\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\)
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(25^{\circ}\)
4 \(54.74^{\circ}\)
Explanation:
: Given that, \(\mathrm{i}_{\mathrm{p}}=54.74^{\circ}\) According to the Brewster's law - \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\sqrt{2}=\mu\) \(\left(\tan 54.74^{\circ}=\sqrt{2}\right)\) According to the Snell's law - \(\mu=\frac{\sin i}{\sin r}\) \(\sqrt{2}=\frac{\sin i}{\sin r}\) Given that, \(\mathrm{i}=45^{\circ}\) \(\sqrt{2}=\frac{1}{\sqrt{2} \sin r}\) \(\sin r=\frac{1}{2}\) \(r=\sin ^{-1}\left(\frac{1}{2}\right)\) \(r=30^{\circ}\)
