283584
The numerical aperture of a microscope is \(\mathbf{0 . 1 2}\), and the wavelength of light used is \(600 \mathrm{~nm}\). Then its limit of resolution will be nearly -
1 \(0.3 \mu \mathrm{m}\)
2 \(1.2 \mu \mathrm{m}\)
3 \(2.3 \mu \mathrm{m}\)
4 \(3.0 \mu \mathrm{m}\)
Explanation:
: Given, \(\lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}\) Numerical aperture \(=\eta \sin \theta=0.12\) The limit of resolution of a microscope, \(x=\frac{0.61 \lambda}{\eta \sin \theta}\) \(x=\frac{0.61 \times 6 \times 10^{-7}}{0.12}=3 \mu \mathrm{m}\)
BITSAT-2005
WAVE OPTICS
283585
Brewster's angle in terms of refractive index (n) of the medium
1 \(\tan ^{-1} \sqrt{\mathrm{n}}\)
2 \(\sin ^{-1} \mathrm{n}\)
3 \(\sin ^{-1} \sqrt{\mathrm{n}}\)
4 \(\tan ^{-1} \mathrm{n}\)
Explanation:
: From Brewster's law, \(\Rightarrow \tan i_p=n\) \(i_p=\tan ^{-1} n\) Where, \(i_P=\) angle of polarization \(\mathrm{n}=\text { refractive index }\)
J and K CET- 2008
WAVE OPTICS
283588
A parallel beam of monochromatic unpolarised light is incident on the transparent dielectric plate of reflective index \(\frac{1}{\sqrt{3}}\). The reflected beam is completely polarised. Then the angle of incidence is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(75^{\circ}\)
Explanation:
Given, refractive index \((\mu)=1 / 3\) According to the Brewster law - \(\tan i_p=\mu\) \(i_p=\tan ^{-1}(\mu)\) \(i_p=\tan ^{-1}(1 / 3)=30^{\circ}\)
AIIMS-2015
WAVE OPTICS
283591
Two polaroids are kept crossed to each other. If one of them is rotated an angle \(60^{\circ}\), the percentage of incident light now transmitted through the system is
1 \(10 \%\)
2 \(20 \%\)
3 \(25 \%\)
4 \(12.5 \%\)
Explanation:
: Let, the intensity of unpolarised light be \(\mathrm{I}_0\), So, the intensity of first polaroid is - \(I_1=\frac{I_0}{2}\) On rotating through \(60^{\circ}\), the intensity of light from second polaroid - \(I_2 =I_1 \cos ^2 60^{\circ}=\left(\frac{I_o}{2}\right)\left(\cos 60^{\circ}\right)^2\) \(=\frac{I_o}{2} \frac{1}{4}=\frac{I_o}{8}=0.125 I_o\)So, percentage of incident light transmitted through the system \(=12.5 \%\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVE OPTICS
283584
The numerical aperture of a microscope is \(\mathbf{0 . 1 2}\), and the wavelength of light used is \(600 \mathrm{~nm}\). Then its limit of resolution will be nearly -
1 \(0.3 \mu \mathrm{m}\)
2 \(1.2 \mu \mathrm{m}\)
3 \(2.3 \mu \mathrm{m}\)
4 \(3.0 \mu \mathrm{m}\)
Explanation:
: Given, \(\lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}\) Numerical aperture \(=\eta \sin \theta=0.12\) The limit of resolution of a microscope, \(x=\frac{0.61 \lambda}{\eta \sin \theta}\) \(x=\frac{0.61 \times 6 \times 10^{-7}}{0.12}=3 \mu \mathrm{m}\)
BITSAT-2005
WAVE OPTICS
283585
Brewster's angle in terms of refractive index (n) of the medium
1 \(\tan ^{-1} \sqrt{\mathrm{n}}\)
2 \(\sin ^{-1} \mathrm{n}\)
3 \(\sin ^{-1} \sqrt{\mathrm{n}}\)
4 \(\tan ^{-1} \mathrm{n}\)
Explanation:
: From Brewster's law, \(\Rightarrow \tan i_p=n\) \(i_p=\tan ^{-1} n\) Where, \(i_P=\) angle of polarization \(\mathrm{n}=\text { refractive index }\)
J and K CET- 2008
WAVE OPTICS
283588
A parallel beam of monochromatic unpolarised light is incident on the transparent dielectric plate of reflective index \(\frac{1}{\sqrt{3}}\). The reflected beam is completely polarised. Then the angle of incidence is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(75^{\circ}\)
Explanation:
Given, refractive index \((\mu)=1 / 3\) According to the Brewster law - \(\tan i_p=\mu\) \(i_p=\tan ^{-1}(\mu)\) \(i_p=\tan ^{-1}(1 / 3)=30^{\circ}\)
AIIMS-2015
WAVE OPTICS
283591
Two polaroids are kept crossed to each other. If one of them is rotated an angle \(60^{\circ}\), the percentage of incident light now transmitted through the system is
1 \(10 \%\)
2 \(20 \%\)
3 \(25 \%\)
4 \(12.5 \%\)
Explanation:
: Let, the intensity of unpolarised light be \(\mathrm{I}_0\), So, the intensity of first polaroid is - \(I_1=\frac{I_0}{2}\) On rotating through \(60^{\circ}\), the intensity of light from second polaroid - \(I_2 =I_1 \cos ^2 60^{\circ}=\left(\frac{I_o}{2}\right)\left(\cos 60^{\circ}\right)^2\) \(=\frac{I_o}{2} \frac{1}{4}=\frac{I_o}{8}=0.125 I_o\)So, percentage of incident light transmitted through the system \(=12.5 \%\).
283584
The numerical aperture of a microscope is \(\mathbf{0 . 1 2}\), and the wavelength of light used is \(600 \mathrm{~nm}\). Then its limit of resolution will be nearly -
1 \(0.3 \mu \mathrm{m}\)
2 \(1.2 \mu \mathrm{m}\)
3 \(2.3 \mu \mathrm{m}\)
4 \(3.0 \mu \mathrm{m}\)
Explanation:
: Given, \(\lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}\) Numerical aperture \(=\eta \sin \theta=0.12\) The limit of resolution of a microscope, \(x=\frac{0.61 \lambda}{\eta \sin \theta}\) \(x=\frac{0.61 \times 6 \times 10^{-7}}{0.12}=3 \mu \mathrm{m}\)
BITSAT-2005
WAVE OPTICS
283585
Brewster's angle in terms of refractive index (n) of the medium
1 \(\tan ^{-1} \sqrt{\mathrm{n}}\)
2 \(\sin ^{-1} \mathrm{n}\)
3 \(\sin ^{-1} \sqrt{\mathrm{n}}\)
4 \(\tan ^{-1} \mathrm{n}\)
Explanation:
: From Brewster's law, \(\Rightarrow \tan i_p=n\) \(i_p=\tan ^{-1} n\) Where, \(i_P=\) angle of polarization \(\mathrm{n}=\text { refractive index }\)
J and K CET- 2008
WAVE OPTICS
283588
A parallel beam of monochromatic unpolarised light is incident on the transparent dielectric plate of reflective index \(\frac{1}{\sqrt{3}}\). The reflected beam is completely polarised. Then the angle of incidence is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(75^{\circ}\)
Explanation:
Given, refractive index \((\mu)=1 / 3\) According to the Brewster law - \(\tan i_p=\mu\) \(i_p=\tan ^{-1}(\mu)\) \(i_p=\tan ^{-1}(1 / 3)=30^{\circ}\)
AIIMS-2015
WAVE OPTICS
283591
Two polaroids are kept crossed to each other. If one of them is rotated an angle \(60^{\circ}\), the percentage of incident light now transmitted through the system is
1 \(10 \%\)
2 \(20 \%\)
3 \(25 \%\)
4 \(12.5 \%\)
Explanation:
: Let, the intensity of unpolarised light be \(\mathrm{I}_0\), So, the intensity of first polaroid is - \(I_1=\frac{I_0}{2}\) On rotating through \(60^{\circ}\), the intensity of light from second polaroid - \(I_2 =I_1 \cos ^2 60^{\circ}=\left(\frac{I_o}{2}\right)\left(\cos 60^{\circ}\right)^2\) \(=\frac{I_o}{2} \frac{1}{4}=\frac{I_o}{8}=0.125 I_o\)So, percentage of incident light transmitted through the system \(=12.5 \%\).
283584
The numerical aperture of a microscope is \(\mathbf{0 . 1 2}\), and the wavelength of light used is \(600 \mathrm{~nm}\). Then its limit of resolution will be nearly -
1 \(0.3 \mu \mathrm{m}\)
2 \(1.2 \mu \mathrm{m}\)
3 \(2.3 \mu \mathrm{m}\)
4 \(3.0 \mu \mathrm{m}\)
Explanation:
: Given, \(\lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}\) Numerical aperture \(=\eta \sin \theta=0.12\) The limit of resolution of a microscope, \(x=\frac{0.61 \lambda}{\eta \sin \theta}\) \(x=\frac{0.61 \times 6 \times 10^{-7}}{0.12}=3 \mu \mathrm{m}\)
BITSAT-2005
WAVE OPTICS
283585
Brewster's angle in terms of refractive index (n) of the medium
1 \(\tan ^{-1} \sqrt{\mathrm{n}}\)
2 \(\sin ^{-1} \mathrm{n}\)
3 \(\sin ^{-1} \sqrt{\mathrm{n}}\)
4 \(\tan ^{-1} \mathrm{n}\)
Explanation:
: From Brewster's law, \(\Rightarrow \tan i_p=n\) \(i_p=\tan ^{-1} n\) Where, \(i_P=\) angle of polarization \(\mathrm{n}=\text { refractive index }\)
J and K CET- 2008
WAVE OPTICS
283588
A parallel beam of monochromatic unpolarised light is incident on the transparent dielectric plate of reflective index \(\frac{1}{\sqrt{3}}\). The reflected beam is completely polarised. Then the angle of incidence is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(75^{\circ}\)
Explanation:
Given, refractive index \((\mu)=1 / 3\) According to the Brewster law - \(\tan i_p=\mu\) \(i_p=\tan ^{-1}(\mu)\) \(i_p=\tan ^{-1}(1 / 3)=30^{\circ}\)
AIIMS-2015
WAVE OPTICS
283591
Two polaroids are kept crossed to each other. If one of them is rotated an angle \(60^{\circ}\), the percentage of incident light now transmitted through the system is
1 \(10 \%\)
2 \(20 \%\)
3 \(25 \%\)
4 \(12.5 \%\)
Explanation:
: Let, the intensity of unpolarised light be \(\mathrm{I}_0\), So, the intensity of first polaroid is - \(I_1=\frac{I_0}{2}\) On rotating through \(60^{\circ}\), the intensity of light from second polaroid - \(I_2 =I_1 \cos ^2 60^{\circ}=\left(\frac{I_o}{2}\right)\left(\cos 60^{\circ}\right)^2\) \(=\frac{I_o}{2} \frac{1}{4}=\frac{I_o}{8}=0.125 I_o\)So, percentage of incident light transmitted through the system \(=12.5 \%\).