282225
If the critical angle for total internal reflection from a medium to vacuum is \({45^{\circ}}^{\circ}\), then velocity of light in the medium is
B: Given, critical angle \(\left(\theta_c\right)=45^{\circ}\)
We know that, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\sin 45^{\circ}=\frac{1}{\mu} \\
\mu=\sqrt{2}
\end{aligned}\)
Therefore, \(\mu=\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in medium }(\mathrm{v})}\)
Putting these value, we get-
\(\begin{aligned}
\sqrt{2}=\frac{3 \times 10^8}{\mathrm{v}} \\
\mathrm{v}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
NEET(Oct.)- 2020
Ray Optics
282226
Transmission of light in optical fibre is due to
1 scattering
2 diffraction
3 polarisation
4 multiple total internal reflections
Explanation:
D: The optical fiber consists of two layer. The outer layer is called cladding (rarer), which has a less refractive index than the inner layer which is called core (denser). The light or signal is sent at an angle greater than the critical angle, so there will be total internal reflection. The signal is sent by multiple internal reflections.
UPSEE 2020
Ray Optics
282267
A vessel of depth \(2 \mathrm{~d} \mathrm{~cm}\) is half filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel seen perpendicularly is -
B: Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
\(=\mathrm{d}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
Ray Optics
282237
In vaccum, light takes time ' \(t\) ' to travel a distance ' \(d\) ' and it takes time ' \(T\) ' to travel a distance ' \(5 \mathrm{~d}\) ' in a denser medium. The critical angle of the given pair of media is
1 \(\sin ^{-1}\left(\frac{t}{T}\right)\)
2 \(\sin ^{-1}\left(\frac{3 t}{T}\right)\)
3 \(\sin ^{-1}\left(\frac{2 t}{T}\right)\)
4 \(\sin ^{-1}\left(\frac{5 t}{T}\right)\)
Explanation:
D: We know that,
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
Also, \(\quad \mu=\frac{\mathrm{c}}{\mathrm{V}}=\frac{\mathrm{d} / \mathrm{t}}{5 \mathrm{~d} / \mathrm{T}}=\frac{\mathrm{T}}{5 \mathrm{t}}\)
Putting the value of \(\mu\) in equation (i), we get -
\(\therefore \quad \theta_c=\sin ^{-1}\left(\frac{5 t}{\mathrm{~T}}\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282225
If the critical angle for total internal reflection from a medium to vacuum is \({45^{\circ}}^{\circ}\), then velocity of light in the medium is
B: Given, critical angle \(\left(\theta_c\right)=45^{\circ}\)
We know that, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\sin 45^{\circ}=\frac{1}{\mu} \\
\mu=\sqrt{2}
\end{aligned}\)
Therefore, \(\mu=\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in medium }(\mathrm{v})}\)
Putting these value, we get-
\(\begin{aligned}
\sqrt{2}=\frac{3 \times 10^8}{\mathrm{v}} \\
\mathrm{v}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
NEET(Oct.)- 2020
Ray Optics
282226
Transmission of light in optical fibre is due to
1 scattering
2 diffraction
3 polarisation
4 multiple total internal reflections
Explanation:
D: The optical fiber consists of two layer. The outer layer is called cladding (rarer), which has a less refractive index than the inner layer which is called core (denser). The light or signal is sent at an angle greater than the critical angle, so there will be total internal reflection. The signal is sent by multiple internal reflections.
UPSEE 2020
Ray Optics
282267
A vessel of depth \(2 \mathrm{~d} \mathrm{~cm}\) is half filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel seen perpendicularly is -
B: Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
\(=\mathrm{d}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
Ray Optics
282237
In vaccum, light takes time ' \(t\) ' to travel a distance ' \(d\) ' and it takes time ' \(T\) ' to travel a distance ' \(5 \mathrm{~d}\) ' in a denser medium. The critical angle of the given pair of media is
1 \(\sin ^{-1}\left(\frac{t}{T}\right)\)
2 \(\sin ^{-1}\left(\frac{3 t}{T}\right)\)
3 \(\sin ^{-1}\left(\frac{2 t}{T}\right)\)
4 \(\sin ^{-1}\left(\frac{5 t}{T}\right)\)
Explanation:
D: We know that,
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
Also, \(\quad \mu=\frac{\mathrm{c}}{\mathrm{V}}=\frac{\mathrm{d} / \mathrm{t}}{5 \mathrm{~d} / \mathrm{T}}=\frac{\mathrm{T}}{5 \mathrm{t}}\)
Putting the value of \(\mu\) in equation (i), we get -
\(\therefore \quad \theta_c=\sin ^{-1}\left(\frac{5 t}{\mathrm{~T}}\right)\)
282225
If the critical angle for total internal reflection from a medium to vacuum is \({45^{\circ}}^{\circ}\), then velocity of light in the medium is
B: Given, critical angle \(\left(\theta_c\right)=45^{\circ}\)
We know that, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\sin 45^{\circ}=\frac{1}{\mu} \\
\mu=\sqrt{2}
\end{aligned}\)
Therefore, \(\mu=\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in medium }(\mathrm{v})}\)
Putting these value, we get-
\(\begin{aligned}
\sqrt{2}=\frac{3 \times 10^8}{\mathrm{v}} \\
\mathrm{v}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
NEET(Oct.)- 2020
Ray Optics
282226
Transmission of light in optical fibre is due to
1 scattering
2 diffraction
3 polarisation
4 multiple total internal reflections
Explanation:
D: The optical fiber consists of two layer. The outer layer is called cladding (rarer), which has a less refractive index than the inner layer which is called core (denser). The light or signal is sent at an angle greater than the critical angle, so there will be total internal reflection. The signal is sent by multiple internal reflections.
UPSEE 2020
Ray Optics
282267
A vessel of depth \(2 \mathrm{~d} \mathrm{~cm}\) is half filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel seen perpendicularly is -
B: Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
\(=\mathrm{d}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
Ray Optics
282237
In vaccum, light takes time ' \(t\) ' to travel a distance ' \(d\) ' and it takes time ' \(T\) ' to travel a distance ' \(5 \mathrm{~d}\) ' in a denser medium. The critical angle of the given pair of media is
1 \(\sin ^{-1}\left(\frac{t}{T}\right)\)
2 \(\sin ^{-1}\left(\frac{3 t}{T}\right)\)
3 \(\sin ^{-1}\left(\frac{2 t}{T}\right)\)
4 \(\sin ^{-1}\left(\frac{5 t}{T}\right)\)
Explanation:
D: We know that,
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
Also, \(\quad \mu=\frac{\mathrm{c}}{\mathrm{V}}=\frac{\mathrm{d} / \mathrm{t}}{5 \mathrm{~d} / \mathrm{T}}=\frac{\mathrm{T}}{5 \mathrm{t}}\)
Putting the value of \(\mu\) in equation (i), we get -
\(\therefore \quad \theta_c=\sin ^{-1}\left(\frac{5 t}{\mathrm{~T}}\right)\)
282225
If the critical angle for total internal reflection from a medium to vacuum is \({45^{\circ}}^{\circ}\), then velocity of light in the medium is
B: Given, critical angle \(\left(\theta_c\right)=45^{\circ}\)
We know that, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\sin 45^{\circ}=\frac{1}{\mu} \\
\mu=\sqrt{2}
\end{aligned}\)
Therefore, \(\mu=\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in medium }(\mathrm{v})}\)
Putting these value, we get-
\(\begin{aligned}
\sqrt{2}=\frac{3 \times 10^8}{\mathrm{v}} \\
\mathrm{v}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
NEET(Oct.)- 2020
Ray Optics
282226
Transmission of light in optical fibre is due to
1 scattering
2 diffraction
3 polarisation
4 multiple total internal reflections
Explanation:
D: The optical fiber consists of two layer. The outer layer is called cladding (rarer), which has a less refractive index than the inner layer which is called core (denser). The light or signal is sent at an angle greater than the critical angle, so there will be total internal reflection. The signal is sent by multiple internal reflections.
UPSEE 2020
Ray Optics
282267
A vessel of depth \(2 \mathrm{~d} \mathrm{~cm}\) is half filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel seen perpendicularly is -
B: Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
\(=\mathrm{d}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)
Ray Optics
282237
In vaccum, light takes time ' \(t\) ' to travel a distance ' \(d\) ' and it takes time ' \(T\) ' to travel a distance ' \(5 \mathrm{~d}\) ' in a denser medium. The critical angle of the given pair of media is
1 \(\sin ^{-1}\left(\frac{t}{T}\right)\)
2 \(\sin ^{-1}\left(\frac{3 t}{T}\right)\)
3 \(\sin ^{-1}\left(\frac{2 t}{T}\right)\)
4 \(\sin ^{-1}\left(\frac{5 t}{T}\right)\)
Explanation:
D: We know that,
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
Also, \(\quad \mu=\frac{\mathrm{c}}{\mathrm{V}}=\frac{\mathrm{d} / \mathrm{t}}{5 \mathrm{~d} / \mathrm{T}}=\frac{\mathrm{T}}{5 \mathrm{t}}\)
Putting the value of \(\mu\) in equation (i), we get -
\(\therefore \quad \theta_c=\sin ^{-1}\left(\frac{5 t}{\mathrm{~T}}\right)\)
