283485
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 \(\frac{\mathrm{I}_0}{2}\)
3 \(\frac{\mathrm{I}_0}{4}\)
4 \(\frac{\mathrm{I}_0}{8}\)
Explanation:
: Unpolarised beam of intensity \(\mathrm{I}_0\). After polarization through first polarizer \((\mathrm{I})=\mathrm{I}_0 \cos ^2 \theta\) Then, after passing from nicol the intensity will be reduced by \(50 \%\). Now, \(I_1=\frac{I_0}{2}\) (Transmitting intensity), \(\left(\mathrm{I}_2\right)=\mathrm{I}_1 \cos ^2\left(60^{\circ}\right)\) \(\frac{I_0}{2} \cos ^2 60^{\circ}=\frac{I_0}{2} \cdot\left(\frac{1}{4}\right)=\frac{I_0}{8}\)
JCECE-2009
WAVE OPTICS
283487
A ray of light is incident on a surface of a plate of glass of refractive index 1.5 at polarising angle. The angle of refraction of the ray will be:
1 \(53.7^{\circ}\)
2 \(43.7^{\circ}\)
3 \(33.7^{\circ}\)
4 none of these
Explanation:
: Given that, \(\mu=1.5=\tan \mathrm{i}_{\mathrm{p}}\) We know that, \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\therefore \quad \mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.5)\) \(=56.18^{\circ}\) Angle of refraction \(\because \quad \mathrm{r} =90^{\circ}-56.8^{\circ}\) \(\mathrm{r} =33.42^{\circ}\)So, the angle of refraction of the ray will be \(33.42^{\circ}\).
JCECE-2003
WAVE OPTICS
283488
A plane polarized light is incident normally on a tourmaline plate. Its \(E\) vectors make an angle of \(60^{\circ}\) with the optic axis of the plate. Find the percentage difference between intial and final intensities.
1 \(90 \%\)
2 \(50 \%\)
3 \(75 \%\)
4 \(25 \%\)
Explanation:
: Given that, \(\theta=60^{\circ}\) Let, the initial intensity of the light \(=I_0\) Final intensity of the light \((\mathrm{I})=\mathrm{I}_{\mathrm{O}} \cos ^2 \theta\) \(\mathrm{I}=\mathrm{I}_{\mathrm{o}} \times \cos ^2 60^{\circ}=0.25 \mathrm{I}_{\mathrm{o}}\) Change in intensity percentage \(=\frac{I_o-I}{I_0} \times 100\) \(=\frac{I_o-0.25 I_o}{I_o} \times 100\) \(=75 \%\)
GUJCET 2016
WAVE OPTICS
283489
de-Broglie wave length of atom at \(\mathrm{T} \mathrm{K}\) absolute temperature will be
1 \(\frac{\mathrm{h}}{\mathrm{mKT}}\)
2 \(\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
3 \(\frac{\sqrt{2 \mathrm{mKT}}}{\mathrm{h}}\)
4 \(\sqrt{2 \mathrm{mKT}}\)
Explanation:
:de-Broglie wavelength provide information about wave nature of a matter the relation between the wavelength and the momentum of the particle is - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\) Where, \(\mathrm{h}=\) Planck's constant \(\lambda=\) wavelength \(\mathrm{p}=\) momentum \(\therefore\) Relation between kinetic energy and absolute temperature is \(\mathrm{E}=\frac{3}{2} \mathrm{KT}\) We also know, the relation between the kinetic energy and momentum of the particle \(\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) Equate the equation (i) and (ii), we get - \(\frac{3}{2} \mathrm{KT}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) \(\mathrm{p}^2=3 \mathrm{mKT}\) \(\mathrm{p}=\sqrt{3 \mathrm{mKT}}\) Now, using de - Broglie relation - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)So, \(\quad \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
283485
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 \(\frac{\mathrm{I}_0}{2}\)
3 \(\frac{\mathrm{I}_0}{4}\)
4 \(\frac{\mathrm{I}_0}{8}\)
Explanation:
: Unpolarised beam of intensity \(\mathrm{I}_0\). After polarization through first polarizer \((\mathrm{I})=\mathrm{I}_0 \cos ^2 \theta\) Then, after passing from nicol the intensity will be reduced by \(50 \%\). Now, \(I_1=\frac{I_0}{2}\) (Transmitting intensity), \(\left(\mathrm{I}_2\right)=\mathrm{I}_1 \cos ^2\left(60^{\circ}\right)\) \(\frac{I_0}{2} \cos ^2 60^{\circ}=\frac{I_0}{2} \cdot\left(\frac{1}{4}\right)=\frac{I_0}{8}\)
JCECE-2009
WAVE OPTICS
283487
A ray of light is incident on a surface of a plate of glass of refractive index 1.5 at polarising angle. The angle of refraction of the ray will be:
1 \(53.7^{\circ}\)
2 \(43.7^{\circ}\)
3 \(33.7^{\circ}\)
4 none of these
Explanation:
: Given that, \(\mu=1.5=\tan \mathrm{i}_{\mathrm{p}}\) We know that, \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\therefore \quad \mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.5)\) \(=56.18^{\circ}\) Angle of refraction \(\because \quad \mathrm{r} =90^{\circ}-56.8^{\circ}\) \(\mathrm{r} =33.42^{\circ}\)So, the angle of refraction of the ray will be \(33.42^{\circ}\).
JCECE-2003
WAVE OPTICS
283488
A plane polarized light is incident normally on a tourmaline plate. Its \(E\) vectors make an angle of \(60^{\circ}\) with the optic axis of the plate. Find the percentage difference between intial and final intensities.
1 \(90 \%\)
2 \(50 \%\)
3 \(75 \%\)
4 \(25 \%\)
Explanation:
: Given that, \(\theta=60^{\circ}\) Let, the initial intensity of the light \(=I_0\) Final intensity of the light \((\mathrm{I})=\mathrm{I}_{\mathrm{O}} \cos ^2 \theta\) \(\mathrm{I}=\mathrm{I}_{\mathrm{o}} \times \cos ^2 60^{\circ}=0.25 \mathrm{I}_{\mathrm{o}}\) Change in intensity percentage \(=\frac{I_o-I}{I_0} \times 100\) \(=\frac{I_o-0.25 I_o}{I_o} \times 100\) \(=75 \%\)
GUJCET 2016
WAVE OPTICS
283489
de-Broglie wave length of atom at \(\mathrm{T} \mathrm{K}\) absolute temperature will be
1 \(\frac{\mathrm{h}}{\mathrm{mKT}}\)
2 \(\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
3 \(\frac{\sqrt{2 \mathrm{mKT}}}{\mathrm{h}}\)
4 \(\sqrt{2 \mathrm{mKT}}\)
Explanation:
:de-Broglie wavelength provide information about wave nature of a matter the relation between the wavelength and the momentum of the particle is - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\) Where, \(\mathrm{h}=\) Planck's constant \(\lambda=\) wavelength \(\mathrm{p}=\) momentum \(\therefore\) Relation between kinetic energy and absolute temperature is \(\mathrm{E}=\frac{3}{2} \mathrm{KT}\) We also know, the relation between the kinetic energy and momentum of the particle \(\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) Equate the equation (i) and (ii), we get - \(\frac{3}{2} \mathrm{KT}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) \(\mathrm{p}^2=3 \mathrm{mKT}\) \(\mathrm{p}=\sqrt{3 \mathrm{mKT}}\) Now, using de - Broglie relation - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)So, \(\quad \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
283485
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 \(\frac{\mathrm{I}_0}{2}\)
3 \(\frac{\mathrm{I}_0}{4}\)
4 \(\frac{\mathrm{I}_0}{8}\)
Explanation:
: Unpolarised beam of intensity \(\mathrm{I}_0\). After polarization through first polarizer \((\mathrm{I})=\mathrm{I}_0 \cos ^2 \theta\) Then, after passing from nicol the intensity will be reduced by \(50 \%\). Now, \(I_1=\frac{I_0}{2}\) (Transmitting intensity), \(\left(\mathrm{I}_2\right)=\mathrm{I}_1 \cos ^2\left(60^{\circ}\right)\) \(\frac{I_0}{2} \cos ^2 60^{\circ}=\frac{I_0}{2} \cdot\left(\frac{1}{4}\right)=\frac{I_0}{8}\)
JCECE-2009
WAVE OPTICS
283487
A ray of light is incident on a surface of a plate of glass of refractive index 1.5 at polarising angle. The angle of refraction of the ray will be:
1 \(53.7^{\circ}\)
2 \(43.7^{\circ}\)
3 \(33.7^{\circ}\)
4 none of these
Explanation:
: Given that, \(\mu=1.5=\tan \mathrm{i}_{\mathrm{p}}\) We know that, \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\therefore \quad \mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.5)\) \(=56.18^{\circ}\) Angle of refraction \(\because \quad \mathrm{r} =90^{\circ}-56.8^{\circ}\) \(\mathrm{r} =33.42^{\circ}\)So, the angle of refraction of the ray will be \(33.42^{\circ}\).
JCECE-2003
WAVE OPTICS
283488
A plane polarized light is incident normally on a tourmaline plate. Its \(E\) vectors make an angle of \(60^{\circ}\) with the optic axis of the plate. Find the percentage difference between intial and final intensities.
1 \(90 \%\)
2 \(50 \%\)
3 \(75 \%\)
4 \(25 \%\)
Explanation:
: Given that, \(\theta=60^{\circ}\) Let, the initial intensity of the light \(=I_0\) Final intensity of the light \((\mathrm{I})=\mathrm{I}_{\mathrm{O}} \cos ^2 \theta\) \(\mathrm{I}=\mathrm{I}_{\mathrm{o}} \times \cos ^2 60^{\circ}=0.25 \mathrm{I}_{\mathrm{o}}\) Change in intensity percentage \(=\frac{I_o-I}{I_0} \times 100\) \(=\frac{I_o-0.25 I_o}{I_o} \times 100\) \(=75 \%\)
GUJCET 2016
WAVE OPTICS
283489
de-Broglie wave length of atom at \(\mathrm{T} \mathrm{K}\) absolute temperature will be
1 \(\frac{\mathrm{h}}{\mathrm{mKT}}\)
2 \(\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
3 \(\frac{\sqrt{2 \mathrm{mKT}}}{\mathrm{h}}\)
4 \(\sqrt{2 \mathrm{mKT}}\)
Explanation:
:de-Broglie wavelength provide information about wave nature of a matter the relation between the wavelength and the momentum of the particle is - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\) Where, \(\mathrm{h}=\) Planck's constant \(\lambda=\) wavelength \(\mathrm{p}=\) momentum \(\therefore\) Relation between kinetic energy and absolute temperature is \(\mathrm{E}=\frac{3}{2} \mathrm{KT}\) We also know, the relation between the kinetic energy and momentum of the particle \(\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) Equate the equation (i) and (ii), we get - \(\frac{3}{2} \mathrm{KT}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) \(\mathrm{p}^2=3 \mathrm{mKT}\) \(\mathrm{p}=\sqrt{3 \mathrm{mKT}}\) Now, using de - Broglie relation - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)So, \(\quad \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
283485
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 \(\frac{\mathrm{I}_0}{2}\)
3 \(\frac{\mathrm{I}_0}{4}\)
4 \(\frac{\mathrm{I}_0}{8}\)
Explanation:
: Unpolarised beam of intensity \(\mathrm{I}_0\). After polarization through first polarizer \((\mathrm{I})=\mathrm{I}_0 \cos ^2 \theta\) Then, after passing from nicol the intensity will be reduced by \(50 \%\). Now, \(I_1=\frac{I_0}{2}\) (Transmitting intensity), \(\left(\mathrm{I}_2\right)=\mathrm{I}_1 \cos ^2\left(60^{\circ}\right)\) \(\frac{I_0}{2} \cos ^2 60^{\circ}=\frac{I_0}{2} \cdot\left(\frac{1}{4}\right)=\frac{I_0}{8}\)
JCECE-2009
WAVE OPTICS
283487
A ray of light is incident on a surface of a plate of glass of refractive index 1.5 at polarising angle. The angle of refraction of the ray will be:
1 \(53.7^{\circ}\)
2 \(43.7^{\circ}\)
3 \(33.7^{\circ}\)
4 none of these
Explanation:
: Given that, \(\mu=1.5=\tan \mathrm{i}_{\mathrm{p}}\) We know that, \(\tan \mathrm{i}_{\mathrm{p}}=\mu\) \(\therefore \quad \mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.5)\) \(=56.18^{\circ}\) Angle of refraction \(\because \quad \mathrm{r} =90^{\circ}-56.8^{\circ}\) \(\mathrm{r} =33.42^{\circ}\)So, the angle of refraction of the ray will be \(33.42^{\circ}\).
JCECE-2003
WAVE OPTICS
283488
A plane polarized light is incident normally on a tourmaline plate. Its \(E\) vectors make an angle of \(60^{\circ}\) with the optic axis of the plate. Find the percentage difference between intial and final intensities.
1 \(90 \%\)
2 \(50 \%\)
3 \(75 \%\)
4 \(25 \%\)
Explanation:
: Given that, \(\theta=60^{\circ}\) Let, the initial intensity of the light \(=I_0\) Final intensity of the light \((\mathrm{I})=\mathrm{I}_{\mathrm{O}} \cos ^2 \theta\) \(\mathrm{I}=\mathrm{I}_{\mathrm{o}} \times \cos ^2 60^{\circ}=0.25 \mathrm{I}_{\mathrm{o}}\) Change in intensity percentage \(=\frac{I_o-I}{I_0} \times 100\) \(=\frac{I_o-0.25 I_o}{I_o} \times 100\) \(=75 \%\)
GUJCET 2016
WAVE OPTICS
283489
de-Broglie wave length of atom at \(\mathrm{T} \mathrm{K}\) absolute temperature will be
1 \(\frac{\mathrm{h}}{\mathrm{mKT}}\)
2 \(\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
3 \(\frac{\sqrt{2 \mathrm{mKT}}}{\mathrm{h}}\)
4 \(\sqrt{2 \mathrm{mKT}}\)
Explanation:
:de-Broglie wavelength provide information about wave nature of a matter the relation between the wavelength and the momentum of the particle is - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\) Where, \(\mathrm{h}=\) Planck's constant \(\lambda=\) wavelength \(\mathrm{p}=\) momentum \(\therefore\) Relation between kinetic energy and absolute temperature is \(\mathrm{E}=\frac{3}{2} \mathrm{KT}\) We also know, the relation between the kinetic energy and momentum of the particle \(\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) Equate the equation (i) and (ii), we get - \(\frac{3}{2} \mathrm{KT}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\) \(\mathrm{p}^2=3 \mathrm{mKT}\) \(\mathrm{p}=\sqrt{3 \mathrm{mKT}}\) Now, using de - Broglie relation - \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)So, \(\quad \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}\)
