283623
A ray of light is incident at an angle \(i\) on a glass slab of refractive index \(\mu\). The angle between reflected and refracted light is \(90^{\circ}\). Then, the relationship between \(i\) and \(\mu\) is
: Given, The angle between reflected and refracted light is \(90^{\circ}\). \(i+r=90^{\circ}\) \(r=90^{\circ}-i\) We know that, \(\mu=\frac{\sin i}{\sin r}\) \(\mu=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) \(\mu=\frac{\sin i}{\cos i}\) \(\mu=\tan i\)
WB JEE 2015
WAVE OPTICS
283561
Two rays of light \(A\) and \(B\) are falling on a glass slab at the angles of incidence \(45^{\circ}\) and \(60^{\circ}\). If the reflected ray of \(A\) is partially polarized and that of \(B\) is completely polarized, then the refractive index of glass is
1 1.33
2 1.414
3 1.5
4 1.65
5 1.732
Explanation:
: Given, incidence angle of \(\mathrm{A}\left(\mathrm{i}_{\mathrm{A}}\right)=45^{\circ}\) \& incidence angle of \(\mathrm{B}\left(\mathrm{i}_{\mathrm{B}}\right)=60^{\circ}\) \(\mathrm{B}\) is completely polarized, Then, \(\mu=\tan \mathrm{i}_{\mathrm{B}}\) \(\mu=\tan 60^{\circ}\) \(\mu=\sqrt{3}=1.732\)
Kerala CEE 04.07.2022
WAVE OPTICS
283565
If \(\theta_P\) is the polarizing angle for a glass plate of refractive index \(\mu\) and critical angle \(\theta_{\mathrm{C}}\) then
: If \(\theta_{\mathrm{p}}\) is the polarizing angle and \(\theta_{\mathrm{C}}\) is the critical angle then the relation between the angle is given by \(\tan \theta_p=\frac{1}{\sin \theta_C}\) \(\tan \theta_p \cdot \sin \theta_C=1\)
Kerala CEE 2021
WAVE OPTICS
283577
Light is incident on a polarizer with intensity \(I_0\). A second prism called analyzer is kept at an angle of \(15^0\), from the first polarizer then the intensity of final emergent light will be :
283623
A ray of light is incident at an angle \(i\) on a glass slab of refractive index \(\mu\). The angle between reflected and refracted light is \(90^{\circ}\). Then, the relationship between \(i\) and \(\mu\) is
: Given, The angle between reflected and refracted light is \(90^{\circ}\). \(i+r=90^{\circ}\) \(r=90^{\circ}-i\) We know that, \(\mu=\frac{\sin i}{\sin r}\) \(\mu=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) \(\mu=\frac{\sin i}{\cos i}\) \(\mu=\tan i\)
WB JEE 2015
WAVE OPTICS
283561
Two rays of light \(A\) and \(B\) are falling on a glass slab at the angles of incidence \(45^{\circ}\) and \(60^{\circ}\). If the reflected ray of \(A\) is partially polarized and that of \(B\) is completely polarized, then the refractive index of glass is
1 1.33
2 1.414
3 1.5
4 1.65
5 1.732
Explanation:
: Given, incidence angle of \(\mathrm{A}\left(\mathrm{i}_{\mathrm{A}}\right)=45^{\circ}\) \& incidence angle of \(\mathrm{B}\left(\mathrm{i}_{\mathrm{B}}\right)=60^{\circ}\) \(\mathrm{B}\) is completely polarized, Then, \(\mu=\tan \mathrm{i}_{\mathrm{B}}\) \(\mu=\tan 60^{\circ}\) \(\mu=\sqrt{3}=1.732\)
Kerala CEE 04.07.2022
WAVE OPTICS
283565
If \(\theta_P\) is the polarizing angle for a glass plate of refractive index \(\mu\) and critical angle \(\theta_{\mathrm{C}}\) then
: If \(\theta_{\mathrm{p}}\) is the polarizing angle and \(\theta_{\mathrm{C}}\) is the critical angle then the relation between the angle is given by \(\tan \theta_p=\frac{1}{\sin \theta_C}\) \(\tan \theta_p \cdot \sin \theta_C=1\)
Kerala CEE 2021
WAVE OPTICS
283577
Light is incident on a polarizer with intensity \(I_0\). A second prism called analyzer is kept at an angle of \(15^0\), from the first polarizer then the intensity of final emergent light will be :
283623
A ray of light is incident at an angle \(i\) on a glass slab of refractive index \(\mu\). The angle between reflected and refracted light is \(90^{\circ}\). Then, the relationship between \(i\) and \(\mu\) is
: Given, The angle between reflected and refracted light is \(90^{\circ}\). \(i+r=90^{\circ}\) \(r=90^{\circ}-i\) We know that, \(\mu=\frac{\sin i}{\sin r}\) \(\mu=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) \(\mu=\frac{\sin i}{\cos i}\) \(\mu=\tan i\)
WB JEE 2015
WAVE OPTICS
283561
Two rays of light \(A\) and \(B\) are falling on a glass slab at the angles of incidence \(45^{\circ}\) and \(60^{\circ}\). If the reflected ray of \(A\) is partially polarized and that of \(B\) is completely polarized, then the refractive index of glass is
1 1.33
2 1.414
3 1.5
4 1.65
5 1.732
Explanation:
: Given, incidence angle of \(\mathrm{A}\left(\mathrm{i}_{\mathrm{A}}\right)=45^{\circ}\) \& incidence angle of \(\mathrm{B}\left(\mathrm{i}_{\mathrm{B}}\right)=60^{\circ}\) \(\mathrm{B}\) is completely polarized, Then, \(\mu=\tan \mathrm{i}_{\mathrm{B}}\) \(\mu=\tan 60^{\circ}\) \(\mu=\sqrt{3}=1.732\)
Kerala CEE 04.07.2022
WAVE OPTICS
283565
If \(\theta_P\) is the polarizing angle for a glass plate of refractive index \(\mu\) and critical angle \(\theta_{\mathrm{C}}\) then
: If \(\theta_{\mathrm{p}}\) is the polarizing angle and \(\theta_{\mathrm{C}}\) is the critical angle then the relation between the angle is given by \(\tan \theta_p=\frac{1}{\sin \theta_C}\) \(\tan \theta_p \cdot \sin \theta_C=1\)
Kerala CEE 2021
WAVE OPTICS
283577
Light is incident on a polarizer with intensity \(I_0\). A second prism called analyzer is kept at an angle of \(15^0\), from the first polarizer then the intensity of final emergent light will be :
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVE OPTICS
283623
A ray of light is incident at an angle \(i\) on a glass slab of refractive index \(\mu\). The angle between reflected and refracted light is \(90^{\circ}\). Then, the relationship between \(i\) and \(\mu\) is
: Given, The angle between reflected and refracted light is \(90^{\circ}\). \(i+r=90^{\circ}\) \(r=90^{\circ}-i\) We know that, \(\mu=\frac{\sin i}{\sin r}\) \(\mu=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) \(\mu=\frac{\sin i}{\cos i}\) \(\mu=\tan i\)
WB JEE 2015
WAVE OPTICS
283561
Two rays of light \(A\) and \(B\) are falling on a glass slab at the angles of incidence \(45^{\circ}\) and \(60^{\circ}\). If the reflected ray of \(A\) is partially polarized and that of \(B\) is completely polarized, then the refractive index of glass is
1 1.33
2 1.414
3 1.5
4 1.65
5 1.732
Explanation:
: Given, incidence angle of \(\mathrm{A}\left(\mathrm{i}_{\mathrm{A}}\right)=45^{\circ}\) \& incidence angle of \(\mathrm{B}\left(\mathrm{i}_{\mathrm{B}}\right)=60^{\circ}\) \(\mathrm{B}\) is completely polarized, Then, \(\mu=\tan \mathrm{i}_{\mathrm{B}}\) \(\mu=\tan 60^{\circ}\) \(\mu=\sqrt{3}=1.732\)
Kerala CEE 04.07.2022
WAVE OPTICS
283565
If \(\theta_P\) is the polarizing angle for a glass plate of refractive index \(\mu\) and critical angle \(\theta_{\mathrm{C}}\) then
: If \(\theta_{\mathrm{p}}\) is the polarizing angle and \(\theta_{\mathrm{C}}\) is the critical angle then the relation between the angle is given by \(\tan \theta_p=\frac{1}{\sin \theta_C}\) \(\tan \theta_p \cdot \sin \theta_C=1\)
Kerala CEE 2021
WAVE OPTICS
283577
Light is incident on a polarizer with intensity \(I_0\). A second prism called analyzer is kept at an angle of \(15^0\), from the first polarizer then the intensity of final emergent light will be :
283623
A ray of light is incident at an angle \(i\) on a glass slab of refractive index \(\mu\). The angle between reflected and refracted light is \(90^{\circ}\). Then, the relationship between \(i\) and \(\mu\) is
: Given, The angle between reflected and refracted light is \(90^{\circ}\). \(i+r=90^{\circ}\) \(r=90^{\circ}-i\) We know that, \(\mu=\frac{\sin i}{\sin r}\) \(\mu=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\) \(\mu=\frac{\sin i}{\cos i}\) \(\mu=\tan i\)
WB JEE 2015
WAVE OPTICS
283561
Two rays of light \(A\) and \(B\) are falling on a glass slab at the angles of incidence \(45^{\circ}\) and \(60^{\circ}\). If the reflected ray of \(A\) is partially polarized and that of \(B\) is completely polarized, then the refractive index of glass is
1 1.33
2 1.414
3 1.5
4 1.65
5 1.732
Explanation:
: Given, incidence angle of \(\mathrm{A}\left(\mathrm{i}_{\mathrm{A}}\right)=45^{\circ}\) \& incidence angle of \(\mathrm{B}\left(\mathrm{i}_{\mathrm{B}}\right)=60^{\circ}\) \(\mathrm{B}\) is completely polarized, Then, \(\mu=\tan \mathrm{i}_{\mathrm{B}}\) \(\mu=\tan 60^{\circ}\) \(\mu=\sqrt{3}=1.732\)
Kerala CEE 04.07.2022
WAVE OPTICS
283565
If \(\theta_P\) is the polarizing angle for a glass plate of refractive index \(\mu\) and critical angle \(\theta_{\mathrm{C}}\) then
: If \(\theta_{\mathrm{p}}\) is the polarizing angle and \(\theta_{\mathrm{C}}\) is the critical angle then the relation between the angle is given by \(\tan \theta_p=\frac{1}{\sin \theta_C}\) \(\tan \theta_p \cdot \sin \theta_C=1\)
Kerala CEE 2021
WAVE OPTICS
283577
Light is incident on a polarizer with intensity \(I_0\). A second prism called analyzer is kept at an angle of \(15^0\), from the first polarizer then the intensity of final emergent light will be :