282221
If the refractive index of glass is \(3 / 2\) and that of water is \(4 / 3\), the critical angle for glass water media is
1 \(\sin ^{-1}\left(\frac{9}{8}\right)\)
2 \(\sin ^{-1}\left(\frac{8}{9}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{2}\right)\)
4 \(\sin ^{-1}\left(\frac{4}{5}\right)\)
Explanation:
B: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=3 / 2\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
\({ }_{\mathrm{w}} \mu_{\mathrm{g}}=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}=\frac{3 / 2}{4 / 3}=\frac{9}{8}\)
Now, critical angle
\(\begin{aligned}
\theta_c=\sin ^{-1}\left(\frac{1}{{ }_w \mu_{\mathrm{g}}}\right)=\sin ^{-1}\left(\frac{1}{\frac{9}{8}}\right) \\
\theta_{\mathrm{c}}=\sin ^{-1}\left(\frac{8}{9}\right)
\end{aligned}\)
Assam CEE-2020
Ray Optics
282222
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: We know that,
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
Net apparent depth \(=\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right) \mathrm{d}\).
BITSAT-2020
Ray Optics
282223
A light ray moving in medium-I (of refractive index \(n_1\) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index \(n_2\) of two media medium-II is decreased, then
1 ray will move completely parallel to the interface
2 ray will be still totally internally reflected at interface
3 ray will be totally transmitted into medium-II only if angle of incidence is increased
4 ray will be totally transmitted in medium-II
Explanation:
B: As' \(n_2^{\prime}\) decreases, \(i_c=\sin ^{-1}\left(n_2 / n_1\right)\) also decreases, so condition \(i>i_c\) is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because \(\mathrm{i}>\mathrm{i}_{\mathrm{c}}\).
UPSEE - 2017
Ray Optics
282224
If eye is kept at a depth \(h\) inside the water of refractive index and viewed outside, then the diameter of circle through which the outer objects become visible, will be
1 \(\frac{\mathrm{h}}{\sqrt{\mu^2+1}}\)
2 \(\frac{h}{\sqrt{\mu^2-1}}\)
3 \(\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}\)
4 \(\frac{\mathrm{h}}{\sqrt{2 \mu^2-1}}\)
Explanation:
C: If radius of circle through which other objects become visible is ' \(r\) ' then -
We know -
\(\begin{aligned}
\sin \theta_c=\frac{1}{\mu}=\frac{r}{\sqrt{r^2+h^2}} \\
\Rightarrow \sqrt{r^2+h^2}=\mu r \\
\Rightarrow \mu^2 r^2=r^2+h^2
\end{aligned}\)
\(\begin{aligned}
\Rightarrow \mathrm{r}=\frac{\mathrm{h}}{\sqrt{\mu^2-1}} \\
\text { Diameter, } \mathrm{d}=2 \mathrm{r} \\
\mathrm{d}=\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}
\end{aligned}\)
282221
If the refractive index of glass is \(3 / 2\) and that of water is \(4 / 3\), the critical angle for glass water media is
1 \(\sin ^{-1}\left(\frac{9}{8}\right)\)
2 \(\sin ^{-1}\left(\frac{8}{9}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{2}\right)\)
4 \(\sin ^{-1}\left(\frac{4}{5}\right)\)
Explanation:
B: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=3 / 2\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
\({ }_{\mathrm{w}} \mu_{\mathrm{g}}=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}=\frac{3 / 2}{4 / 3}=\frac{9}{8}\)
Now, critical angle
\(\begin{aligned}
\theta_c=\sin ^{-1}\left(\frac{1}{{ }_w \mu_{\mathrm{g}}}\right)=\sin ^{-1}\left(\frac{1}{\frac{9}{8}}\right) \\
\theta_{\mathrm{c}}=\sin ^{-1}\left(\frac{8}{9}\right)
\end{aligned}\)
Assam CEE-2020
Ray Optics
282222
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: We know that,
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
Net apparent depth \(=\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right) \mathrm{d}\).
BITSAT-2020
Ray Optics
282223
A light ray moving in medium-I (of refractive index \(n_1\) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index \(n_2\) of two media medium-II is decreased, then
1 ray will move completely parallel to the interface
2 ray will be still totally internally reflected at interface
3 ray will be totally transmitted into medium-II only if angle of incidence is increased
4 ray will be totally transmitted in medium-II
Explanation:
B: As' \(n_2^{\prime}\) decreases, \(i_c=\sin ^{-1}\left(n_2 / n_1\right)\) also decreases, so condition \(i>i_c\) is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because \(\mathrm{i}>\mathrm{i}_{\mathrm{c}}\).
UPSEE - 2017
Ray Optics
282224
If eye is kept at a depth \(h\) inside the water of refractive index and viewed outside, then the diameter of circle through which the outer objects become visible, will be
1 \(\frac{\mathrm{h}}{\sqrt{\mu^2+1}}\)
2 \(\frac{h}{\sqrt{\mu^2-1}}\)
3 \(\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}\)
4 \(\frac{\mathrm{h}}{\sqrt{2 \mu^2-1}}\)
Explanation:
C: If radius of circle through which other objects become visible is ' \(r\) ' then -
We know -
\(\begin{aligned}
\sin \theta_c=\frac{1}{\mu}=\frac{r}{\sqrt{r^2+h^2}} \\
\Rightarrow \sqrt{r^2+h^2}=\mu r \\
\Rightarrow \mu^2 r^2=r^2+h^2
\end{aligned}\)
\(\begin{aligned}
\Rightarrow \mathrm{r}=\frac{\mathrm{h}}{\sqrt{\mu^2-1}} \\
\text { Diameter, } \mathrm{d}=2 \mathrm{r} \\
\mathrm{d}=\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}
\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282221
If the refractive index of glass is \(3 / 2\) and that of water is \(4 / 3\), the critical angle for glass water media is
1 \(\sin ^{-1}\left(\frac{9}{8}\right)\)
2 \(\sin ^{-1}\left(\frac{8}{9}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{2}\right)\)
4 \(\sin ^{-1}\left(\frac{4}{5}\right)\)
Explanation:
B: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=3 / 2\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
\({ }_{\mathrm{w}} \mu_{\mathrm{g}}=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}=\frac{3 / 2}{4 / 3}=\frac{9}{8}\)
Now, critical angle
\(\begin{aligned}
\theta_c=\sin ^{-1}\left(\frac{1}{{ }_w \mu_{\mathrm{g}}}\right)=\sin ^{-1}\left(\frac{1}{\frac{9}{8}}\right) \\
\theta_{\mathrm{c}}=\sin ^{-1}\left(\frac{8}{9}\right)
\end{aligned}\)
Assam CEE-2020
Ray Optics
282222
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: We know that,
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
Net apparent depth \(=\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right) \mathrm{d}\).
BITSAT-2020
Ray Optics
282223
A light ray moving in medium-I (of refractive index \(n_1\) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index \(n_2\) of two media medium-II is decreased, then
1 ray will move completely parallel to the interface
2 ray will be still totally internally reflected at interface
3 ray will be totally transmitted into medium-II only if angle of incidence is increased
4 ray will be totally transmitted in medium-II
Explanation:
B: As' \(n_2^{\prime}\) decreases, \(i_c=\sin ^{-1}\left(n_2 / n_1\right)\) also decreases, so condition \(i>i_c\) is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because \(\mathrm{i}>\mathrm{i}_{\mathrm{c}}\).
UPSEE - 2017
Ray Optics
282224
If eye is kept at a depth \(h\) inside the water of refractive index and viewed outside, then the diameter of circle through which the outer objects become visible, will be
1 \(\frac{\mathrm{h}}{\sqrt{\mu^2+1}}\)
2 \(\frac{h}{\sqrt{\mu^2-1}}\)
3 \(\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}\)
4 \(\frac{\mathrm{h}}{\sqrt{2 \mu^2-1}}\)
Explanation:
C: If radius of circle through which other objects become visible is ' \(r\) ' then -
We know -
\(\begin{aligned}
\sin \theta_c=\frac{1}{\mu}=\frac{r}{\sqrt{r^2+h^2}} \\
\Rightarrow \sqrt{r^2+h^2}=\mu r \\
\Rightarrow \mu^2 r^2=r^2+h^2
\end{aligned}\)
\(\begin{aligned}
\Rightarrow \mathrm{r}=\frac{\mathrm{h}}{\sqrt{\mu^2-1}} \\
\text { Diameter, } \mathrm{d}=2 \mathrm{r} \\
\mathrm{d}=\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}
\end{aligned}\)
282221
If the refractive index of glass is \(3 / 2\) and that of water is \(4 / 3\), the critical angle for glass water media is
1 \(\sin ^{-1}\left(\frac{9}{8}\right)\)
2 \(\sin ^{-1}\left(\frac{8}{9}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{2}\right)\)
4 \(\sin ^{-1}\left(\frac{4}{5}\right)\)
Explanation:
B: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=3 / 2\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=4 / 3\)
\({ }_{\mathrm{w}} \mu_{\mathrm{g}}=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}=\frac{3 / 2}{4 / 3}=\frac{9}{8}\)
Now, critical angle
\(\begin{aligned}
\theta_c=\sin ^{-1}\left(\frac{1}{{ }_w \mu_{\mathrm{g}}}\right)=\sin ^{-1}\left(\frac{1}{\frac{9}{8}}\right) \\
\theta_{\mathrm{c}}=\sin ^{-1}\left(\frac{8}{9}\right)
\end{aligned}\)
Assam CEE-2020
Ray Optics
282222
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: We know that,
\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Net apparent depth \(=\frac{\mathrm{d}}{\mu_1}+\frac{\mathrm{d}}{\mu_2}\)
Net apparent depth \(=\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right) \mathrm{d}\).
BITSAT-2020
Ray Optics
282223
A light ray moving in medium-I (of refractive index \(n_1\) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index \(n_2\) of two media medium-II is decreased, then
1 ray will move completely parallel to the interface
2 ray will be still totally internally reflected at interface
3 ray will be totally transmitted into medium-II only if angle of incidence is increased
4 ray will be totally transmitted in medium-II
Explanation:
B: As' \(n_2^{\prime}\) decreases, \(i_c=\sin ^{-1}\left(n_2 / n_1\right)\) also decreases, so condition \(i>i_c\) is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because \(\mathrm{i}>\mathrm{i}_{\mathrm{c}}\).
UPSEE - 2017
Ray Optics
282224
If eye is kept at a depth \(h\) inside the water of refractive index and viewed outside, then the diameter of circle through which the outer objects become visible, will be
1 \(\frac{\mathrm{h}}{\sqrt{\mu^2+1}}\)
2 \(\frac{h}{\sqrt{\mu^2-1}}\)
3 \(\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}\)
4 \(\frac{\mathrm{h}}{\sqrt{2 \mu^2-1}}\)
Explanation:
C: If radius of circle through which other objects become visible is ' \(r\) ' then -
We know -
\(\begin{aligned}
\sin \theta_c=\frac{1}{\mu}=\frac{r}{\sqrt{r^2+h^2}} \\
\Rightarrow \sqrt{r^2+h^2}=\mu r \\
\Rightarrow \mu^2 r^2=r^2+h^2
\end{aligned}\)
\(\begin{aligned}
\Rightarrow \mathrm{r}=\frac{\mathrm{h}}{\sqrt{\mu^2-1}} \\
\text { Diameter, } \mathrm{d}=2 \mathrm{r} \\
\mathrm{d}=\frac{2 \mathrm{~h}}{\sqrt{\mu^2-1}}
\end{aligned}\)
