282236
The critical angle is maximum when light travels from \(\left({ }_a \mu_w=\frac{4}{3},{ }_a \mu_g=\frac{3}{2}\right)\)
1 air to water
2 glass to water
3 glass to air
4 water to air
Explanation:
B: Given, \({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3} \approx 1.33,{ }_{\mathrm{a}} \mu_{\mathrm{g}}=\frac{3}{2} \approx 1.5\)
Therefore, \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{3 / 2}{4 / 3}=\frac{9}{8} \approx 1.1\)
And \(\quad{ }_{\mathrm{g}} \mu_{\mathrm{w}}=\frac{4 / 3}{3 / 2}=\frac{8}{9} \approx 0.88\)
Now, critical angle \(\left(\theta_c\right) \propto \frac{1}{\mu}\)
Hence, \(\theta_{\mathrm{c}}\) is maximum for minimum refractive index. When light travels from glass to water \(\theta_{\mathrm{c}}\) is maximum.
MHT-CET 2020
Ray Optics
282238
Time taken by sunlight to penetrate \(2 \mathrm{~mm}\) through a glass slab is of the order Refractive index of glass \(=1.5\), velocity of light in air \(=\) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(10^{-19} \mathrm{~s}\)
2 \(10^5 \mathrm{~s}\)
3 \(10^{-11} \mathrm{~s}\)
4 \(10^{-10} \mathrm{~s}\)
Explanation:
C: Given, refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\) Thickness \((\mathrm{d})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
We know that,
\(\text { Time }(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{v}}=\frac{\mathrm{d} \mu}{\mathrm{c}} \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mu}\right)\)
\(\begin{aligned}
t=\frac{2 \times 10^{-3} \times 1.5}{3 \times 10^8} \\
t=10^{-11} s
\end{aligned}\)
MHT-CET 2020
Ray Optics
282239
A ray of light travels from a medium of refractive index 2 to another medium of refractive index \(\sqrt{2}\). Total internal reflection takes place when the angle of incidence is
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(80^{\circ}\)
Explanation:
B: Given, \(\mu_1=2, \mu_2=\sqrt{2}\)
\(\mu={ }_2 \mu_1=\frac{\mu_1}{\mu_2}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
for total internal reflection
critical angle, \(\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\theta_c=45^{\circ}
\end{aligned}\)
AP EAMCET (17.09.2020) Shift-II
Ray Optics
282240
Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because
1 the holes through the fibre are extremely fine
2 light rays are absorbed at the entry end and relieved at the exit end of the fibre.
3 scattering of light occurs inside the fibre.
4 successive total internal reflections occur as a ray moves through the fibre.
Ans : (d): In total internal reflection, when a ray of light travels from denser to rarer medium, it bends away from the normal because the angle of incidence in denser medium increases. The angle of refraction in rarer medium also increases and at a certain angle, the angle of refraction becomes \(90^{\circ}\). This angle of incidence is called critical angle. When the angle of incidence is greater than the critical angle then the light ray comes back into the same medium after reflection from interface. This phenomenon is known as the total internal reflection
Hence, light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because successive total internal reflections occur as a ray moves through the fibre.
282236
The critical angle is maximum when light travels from \(\left({ }_a \mu_w=\frac{4}{3},{ }_a \mu_g=\frac{3}{2}\right)\)
1 air to water
2 glass to water
3 glass to air
4 water to air
Explanation:
B: Given, \({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3} \approx 1.33,{ }_{\mathrm{a}} \mu_{\mathrm{g}}=\frac{3}{2} \approx 1.5\)
Therefore, \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{3 / 2}{4 / 3}=\frac{9}{8} \approx 1.1\)
And \(\quad{ }_{\mathrm{g}} \mu_{\mathrm{w}}=\frac{4 / 3}{3 / 2}=\frac{8}{9} \approx 0.88\)
Now, critical angle \(\left(\theta_c\right) \propto \frac{1}{\mu}\)
Hence, \(\theta_{\mathrm{c}}\) is maximum for minimum refractive index. When light travels from glass to water \(\theta_{\mathrm{c}}\) is maximum.
MHT-CET 2020
Ray Optics
282238
Time taken by sunlight to penetrate \(2 \mathrm{~mm}\) through a glass slab is of the order Refractive index of glass \(=1.5\), velocity of light in air \(=\) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(10^{-19} \mathrm{~s}\)
2 \(10^5 \mathrm{~s}\)
3 \(10^{-11} \mathrm{~s}\)
4 \(10^{-10} \mathrm{~s}\)
Explanation:
C: Given, refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\) Thickness \((\mathrm{d})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
We know that,
\(\text { Time }(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{v}}=\frac{\mathrm{d} \mu}{\mathrm{c}} \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mu}\right)\)
\(\begin{aligned}
t=\frac{2 \times 10^{-3} \times 1.5}{3 \times 10^8} \\
t=10^{-11} s
\end{aligned}\)
MHT-CET 2020
Ray Optics
282239
A ray of light travels from a medium of refractive index 2 to another medium of refractive index \(\sqrt{2}\). Total internal reflection takes place when the angle of incidence is
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(80^{\circ}\)
Explanation:
B: Given, \(\mu_1=2, \mu_2=\sqrt{2}\)
\(\mu={ }_2 \mu_1=\frac{\mu_1}{\mu_2}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
for total internal reflection
critical angle, \(\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\theta_c=45^{\circ}
\end{aligned}\)
AP EAMCET (17.09.2020) Shift-II
Ray Optics
282240
Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because
1 the holes through the fibre are extremely fine
2 light rays are absorbed at the entry end and relieved at the exit end of the fibre.
3 scattering of light occurs inside the fibre.
4 successive total internal reflections occur as a ray moves through the fibre.
Ans : (d): In total internal reflection, when a ray of light travels from denser to rarer medium, it bends away from the normal because the angle of incidence in denser medium increases. The angle of refraction in rarer medium also increases and at a certain angle, the angle of refraction becomes \(90^{\circ}\). This angle of incidence is called critical angle. When the angle of incidence is greater than the critical angle then the light ray comes back into the same medium after reflection from interface. This phenomenon is known as the total internal reflection
Hence, light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because successive total internal reflections occur as a ray moves through the fibre.
282236
The critical angle is maximum when light travels from \(\left({ }_a \mu_w=\frac{4}{3},{ }_a \mu_g=\frac{3}{2}\right)\)
1 air to water
2 glass to water
3 glass to air
4 water to air
Explanation:
B: Given, \({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3} \approx 1.33,{ }_{\mathrm{a}} \mu_{\mathrm{g}}=\frac{3}{2} \approx 1.5\)
Therefore, \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{3 / 2}{4 / 3}=\frac{9}{8} \approx 1.1\)
And \(\quad{ }_{\mathrm{g}} \mu_{\mathrm{w}}=\frac{4 / 3}{3 / 2}=\frac{8}{9} \approx 0.88\)
Now, critical angle \(\left(\theta_c\right) \propto \frac{1}{\mu}\)
Hence, \(\theta_{\mathrm{c}}\) is maximum for minimum refractive index. When light travels from glass to water \(\theta_{\mathrm{c}}\) is maximum.
MHT-CET 2020
Ray Optics
282238
Time taken by sunlight to penetrate \(2 \mathrm{~mm}\) through a glass slab is of the order Refractive index of glass \(=1.5\), velocity of light in air \(=\) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(10^{-19} \mathrm{~s}\)
2 \(10^5 \mathrm{~s}\)
3 \(10^{-11} \mathrm{~s}\)
4 \(10^{-10} \mathrm{~s}\)
Explanation:
C: Given, refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\) Thickness \((\mathrm{d})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
We know that,
\(\text { Time }(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{v}}=\frac{\mathrm{d} \mu}{\mathrm{c}} \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mu}\right)\)
\(\begin{aligned}
t=\frac{2 \times 10^{-3} \times 1.5}{3 \times 10^8} \\
t=10^{-11} s
\end{aligned}\)
MHT-CET 2020
Ray Optics
282239
A ray of light travels from a medium of refractive index 2 to another medium of refractive index \(\sqrt{2}\). Total internal reflection takes place when the angle of incidence is
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(80^{\circ}\)
Explanation:
B: Given, \(\mu_1=2, \mu_2=\sqrt{2}\)
\(\mu={ }_2 \mu_1=\frac{\mu_1}{\mu_2}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
for total internal reflection
critical angle, \(\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\theta_c=45^{\circ}
\end{aligned}\)
AP EAMCET (17.09.2020) Shift-II
Ray Optics
282240
Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because
1 the holes through the fibre are extremely fine
2 light rays are absorbed at the entry end and relieved at the exit end of the fibre.
3 scattering of light occurs inside the fibre.
4 successive total internal reflections occur as a ray moves through the fibre.
Ans : (d): In total internal reflection, when a ray of light travels from denser to rarer medium, it bends away from the normal because the angle of incidence in denser medium increases. The angle of refraction in rarer medium also increases and at a certain angle, the angle of refraction becomes \(90^{\circ}\). This angle of incidence is called critical angle. When the angle of incidence is greater than the critical angle then the light ray comes back into the same medium after reflection from interface. This phenomenon is known as the total internal reflection
Hence, light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because successive total internal reflections occur as a ray moves through the fibre.
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Ray Optics
282236
The critical angle is maximum when light travels from \(\left({ }_a \mu_w=\frac{4}{3},{ }_a \mu_g=\frac{3}{2}\right)\)
1 air to water
2 glass to water
3 glass to air
4 water to air
Explanation:
B: Given, \({ }_{\mathrm{a}} \mu_{\mathrm{w}}=\frac{4}{3} \approx 1.33,{ }_{\mathrm{a}} \mu_{\mathrm{g}}=\frac{3}{2} \approx 1.5\)
Therefore, \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{3 / 2}{4 / 3}=\frac{9}{8} \approx 1.1\)
And \(\quad{ }_{\mathrm{g}} \mu_{\mathrm{w}}=\frac{4 / 3}{3 / 2}=\frac{8}{9} \approx 0.88\)
Now, critical angle \(\left(\theta_c\right) \propto \frac{1}{\mu}\)
Hence, \(\theta_{\mathrm{c}}\) is maximum for minimum refractive index. When light travels from glass to water \(\theta_{\mathrm{c}}\) is maximum.
MHT-CET 2020
Ray Optics
282238
Time taken by sunlight to penetrate \(2 \mathrm{~mm}\) through a glass slab is of the order Refractive index of glass \(=1.5\), velocity of light in air \(=\) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(10^{-19} \mathrm{~s}\)
2 \(10^5 \mathrm{~s}\)
3 \(10^{-11} \mathrm{~s}\)
4 \(10^{-10} \mathrm{~s}\)
Explanation:
C: Given, refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\) Thickness \((\mathrm{d})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
We know that,
\(\text { Time }(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{v}}=\frac{\mathrm{d} \mu}{\mathrm{c}} \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mu}\right)\)
\(\begin{aligned}
t=\frac{2 \times 10^{-3} \times 1.5}{3 \times 10^8} \\
t=10^{-11} s
\end{aligned}\)
MHT-CET 2020
Ray Optics
282239
A ray of light travels from a medium of refractive index 2 to another medium of refractive index \(\sqrt{2}\). Total internal reflection takes place when the angle of incidence is
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(80^{\circ}\)
Explanation:
B: Given, \(\mu_1=2, \mu_2=\sqrt{2}\)
\(\mu={ }_2 \mu_1=\frac{\mu_1}{\mu_2}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
for total internal reflection
critical angle, \(\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\mu}\right)\)
\(\begin{aligned}
\theta_c=\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\theta_c=45^{\circ}
\end{aligned}\)
AP EAMCET (17.09.2020) Shift-II
Ray Optics
282240
Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because
1 the holes through the fibre are extremely fine
2 light rays are absorbed at the entry end and relieved at the exit end of the fibre.
3 scattering of light occurs inside the fibre.
4 successive total internal reflections occur as a ray moves through the fibre.
Ans : (d): In total internal reflection, when a ray of light travels from denser to rarer medium, it bends away from the normal because the angle of incidence in denser medium increases. The angle of refraction in rarer medium also increases and at a certain angle, the angle of refraction becomes \(90^{\circ}\). This angle of incidence is called critical angle. When the angle of incidence is greater than the critical angle then the light ray comes back into the same medium after reflection from interface. This phenomenon is known as the total internal reflection
Hence, light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because successive total internal reflections occur as a ray moves through the fibre.