282778
The angle of a prism is \(60^{\circ}\) and the angle of minimum deviation of light passing through it is observed to be \(40^{\circ}\). The angle of incidence of light is
1 \(30^{\circ}\)
2 \(40^{\circ}\)
3 \(50^{\circ}\)
4 \(60^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}=60^{\circ}\) and \(\delta_{\mathrm{m}}=40^{\circ}\)
We know that,
Angle of incident \((i)=\frac{A+\delta_m}{2}\)
\(\begin{aligned}
i=\frac{60^{\circ}+40^{\circ}}{2} \\
i=50^{\circ}
\end{aligned}\)
AP EAMCET-1993
Ray Optics
282674
Find the value of the angle of emergence from the prism. Refractive index of the glass is \(\sqrt{3}\).
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A: Given, Refractive index of glass \((\mu)=\sqrt{3}\)
From figure -
\(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}=30^{\circ}\)
\(\left(\because \mathrm{r}_1=0^{\circ}\right)\)
\(r_2=30^{\circ}\)
From Snell's law \(\mu_1 \sin r_2=\mu_0\) sin \(\mathrm{e}\) \(\sqrt{3} \times \sin 30^{\circ}=1 \times \sin \mathrm{e}\)
\(e=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) e=60^{\circ}\)
NEET- 2021
Ray Optics
282675
A prism of refractive index \(\sqrt{2}\) has a refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers a minimum deviation?
1 \(45^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
A: Given, \(\mu=\sqrt{2}\)
Prism angle \((A)=60^{\circ}\)
For minimum deviation -
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For minimum deviation angle of incidence is equal to angle of emergence
Hence, \(i=\frac{A+\delta_m}{2}\)
Then, \(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \frac{60^{\circ}}{2}}\)
\(\sin \mathrm{i}=\sqrt{2} \times \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}\)
\(\sin \mathrm{i}=\frac{1}{\sqrt{2}}\)
\(\mathrm{i}=45^{\circ}\)
TRIPURA-2021
Ray Optics
282711
The angle of minimum deviation for a prism of angle \(A\) is \(180^{\circ}-2 A\). The refractive index is
1 \(\sin \frac{A}{2}\)
2 \(\cos \frac{\mathrm{A}}{2}\)
3 \(\tan \frac{\mathrm{A}}{2}\)
4 \(\cot \frac{\mathrm{A}}{2}\)
Explanation:
D: Given, \(\delta_{\mathrm{m}}=180^{\circ}-2 \mathrm{~A}\)
We know,
\(\begin{aligned}
\text { Refractive index }(\mu)=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)} \\
=\frac{\sin \left(\frac{\mathrm{A}+180^{\circ}-2 \mathrm{~A}}{2}\right)}{\sin \mathrm{A} / 2} \\
\mu=\cot \mathrm{A} / 2 \\
\end{aligned}\)
Manipal UGET-2017
Ray Optics
282676
A thin prism of angle \(6^{\circ}\) made up of glass of refractive index 1.5 is combined with another prism made up of glass of refractive index 1.75 to produce dispersion without deviation. Then find the angle of the second prism.
1 \(7^{\circ}\)
2 \(9^{\circ}\)
3 \(4^{\circ}\)
4 \(5^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}_1=6^{\circ}, \mu_1=1.5\)
\(\mathrm{A}_2=?, \quad \mu_2=1.75\)
We know that \(\delta=(\mu-1) \mathrm{A}\)
Given here \(\quad \delta_1=\delta_2\)
\(\begin{aligned}
\mathrm{A}_1\left(\mu_1-1\right)=\mathrm{A}_2\left(\mu_2-1\right) \\
6^{\mathrm{o}}(1.5-1)=\mathrm{A}_2(1.75-1) \\
\mathrm{A}_2=\frac{6^{\mathrm{o}} \times 0.5}{0.75} \\
\mathrm{~A}_2=4^{\mathrm{o}}
\end{aligned}\)
282778
The angle of a prism is \(60^{\circ}\) and the angle of minimum deviation of light passing through it is observed to be \(40^{\circ}\). The angle of incidence of light is
1 \(30^{\circ}\)
2 \(40^{\circ}\)
3 \(50^{\circ}\)
4 \(60^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}=60^{\circ}\) and \(\delta_{\mathrm{m}}=40^{\circ}\)
We know that,
Angle of incident \((i)=\frac{A+\delta_m}{2}\)
\(\begin{aligned}
i=\frac{60^{\circ}+40^{\circ}}{2} \\
i=50^{\circ}
\end{aligned}\)
AP EAMCET-1993
Ray Optics
282674
Find the value of the angle of emergence from the prism. Refractive index of the glass is \(\sqrt{3}\).
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A: Given, Refractive index of glass \((\mu)=\sqrt{3}\)
From figure -
\(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}=30^{\circ}\)
\(\left(\because \mathrm{r}_1=0^{\circ}\right)\)
\(r_2=30^{\circ}\)
From Snell's law \(\mu_1 \sin r_2=\mu_0\) sin \(\mathrm{e}\) \(\sqrt{3} \times \sin 30^{\circ}=1 \times \sin \mathrm{e}\)
\(e=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) e=60^{\circ}\)
NEET- 2021
Ray Optics
282675
A prism of refractive index \(\sqrt{2}\) has a refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers a minimum deviation?
1 \(45^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
A: Given, \(\mu=\sqrt{2}\)
Prism angle \((A)=60^{\circ}\)
For minimum deviation -
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For minimum deviation angle of incidence is equal to angle of emergence
Hence, \(i=\frac{A+\delta_m}{2}\)
Then, \(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \frac{60^{\circ}}{2}}\)
\(\sin \mathrm{i}=\sqrt{2} \times \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}\)
\(\sin \mathrm{i}=\frac{1}{\sqrt{2}}\)
\(\mathrm{i}=45^{\circ}\)
TRIPURA-2021
Ray Optics
282711
The angle of minimum deviation for a prism of angle \(A\) is \(180^{\circ}-2 A\). The refractive index is
1 \(\sin \frac{A}{2}\)
2 \(\cos \frac{\mathrm{A}}{2}\)
3 \(\tan \frac{\mathrm{A}}{2}\)
4 \(\cot \frac{\mathrm{A}}{2}\)
Explanation:
D: Given, \(\delta_{\mathrm{m}}=180^{\circ}-2 \mathrm{~A}\)
We know,
\(\begin{aligned}
\text { Refractive index }(\mu)=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)} \\
=\frac{\sin \left(\frac{\mathrm{A}+180^{\circ}-2 \mathrm{~A}}{2}\right)}{\sin \mathrm{A} / 2} \\
\mu=\cot \mathrm{A} / 2 \\
\end{aligned}\)
Manipal UGET-2017
Ray Optics
282676
A thin prism of angle \(6^{\circ}\) made up of glass of refractive index 1.5 is combined with another prism made up of glass of refractive index 1.75 to produce dispersion without deviation. Then find the angle of the second prism.
1 \(7^{\circ}\)
2 \(9^{\circ}\)
3 \(4^{\circ}\)
4 \(5^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}_1=6^{\circ}, \mu_1=1.5\)
\(\mathrm{A}_2=?, \quad \mu_2=1.75\)
We know that \(\delta=(\mu-1) \mathrm{A}\)
Given here \(\quad \delta_1=\delta_2\)
\(\begin{aligned}
\mathrm{A}_1\left(\mu_1-1\right)=\mathrm{A}_2\left(\mu_2-1\right) \\
6^{\mathrm{o}}(1.5-1)=\mathrm{A}_2(1.75-1) \\
\mathrm{A}_2=\frac{6^{\mathrm{o}} \times 0.5}{0.75} \\
\mathrm{~A}_2=4^{\mathrm{o}}
\end{aligned}\)
282778
The angle of a prism is \(60^{\circ}\) and the angle of minimum deviation of light passing through it is observed to be \(40^{\circ}\). The angle of incidence of light is
1 \(30^{\circ}\)
2 \(40^{\circ}\)
3 \(50^{\circ}\)
4 \(60^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}=60^{\circ}\) and \(\delta_{\mathrm{m}}=40^{\circ}\)
We know that,
Angle of incident \((i)=\frac{A+\delta_m}{2}\)
\(\begin{aligned}
i=\frac{60^{\circ}+40^{\circ}}{2} \\
i=50^{\circ}
\end{aligned}\)
AP EAMCET-1993
Ray Optics
282674
Find the value of the angle of emergence from the prism. Refractive index of the glass is \(\sqrt{3}\).
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A: Given, Refractive index of glass \((\mu)=\sqrt{3}\)
From figure -
\(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}=30^{\circ}\)
\(\left(\because \mathrm{r}_1=0^{\circ}\right)\)
\(r_2=30^{\circ}\)
From Snell's law \(\mu_1 \sin r_2=\mu_0\) sin \(\mathrm{e}\) \(\sqrt{3} \times \sin 30^{\circ}=1 \times \sin \mathrm{e}\)
\(e=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) e=60^{\circ}\)
NEET- 2021
Ray Optics
282675
A prism of refractive index \(\sqrt{2}\) has a refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers a minimum deviation?
1 \(45^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
A: Given, \(\mu=\sqrt{2}\)
Prism angle \((A)=60^{\circ}\)
For minimum deviation -
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For minimum deviation angle of incidence is equal to angle of emergence
Hence, \(i=\frac{A+\delta_m}{2}\)
Then, \(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \frac{60^{\circ}}{2}}\)
\(\sin \mathrm{i}=\sqrt{2} \times \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}\)
\(\sin \mathrm{i}=\frac{1}{\sqrt{2}}\)
\(\mathrm{i}=45^{\circ}\)
TRIPURA-2021
Ray Optics
282711
The angle of minimum deviation for a prism of angle \(A\) is \(180^{\circ}-2 A\). The refractive index is
1 \(\sin \frac{A}{2}\)
2 \(\cos \frac{\mathrm{A}}{2}\)
3 \(\tan \frac{\mathrm{A}}{2}\)
4 \(\cot \frac{\mathrm{A}}{2}\)
Explanation:
D: Given, \(\delta_{\mathrm{m}}=180^{\circ}-2 \mathrm{~A}\)
We know,
\(\begin{aligned}
\text { Refractive index }(\mu)=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)} \\
=\frac{\sin \left(\frac{\mathrm{A}+180^{\circ}-2 \mathrm{~A}}{2}\right)}{\sin \mathrm{A} / 2} \\
\mu=\cot \mathrm{A} / 2 \\
\end{aligned}\)
Manipal UGET-2017
Ray Optics
282676
A thin prism of angle \(6^{\circ}\) made up of glass of refractive index 1.5 is combined with another prism made up of glass of refractive index 1.75 to produce dispersion without deviation. Then find the angle of the second prism.
1 \(7^{\circ}\)
2 \(9^{\circ}\)
3 \(4^{\circ}\)
4 \(5^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}_1=6^{\circ}, \mu_1=1.5\)
\(\mathrm{A}_2=?, \quad \mu_2=1.75\)
We know that \(\delta=(\mu-1) \mathrm{A}\)
Given here \(\quad \delta_1=\delta_2\)
\(\begin{aligned}
\mathrm{A}_1\left(\mu_1-1\right)=\mathrm{A}_2\left(\mu_2-1\right) \\
6^{\mathrm{o}}(1.5-1)=\mathrm{A}_2(1.75-1) \\
\mathrm{A}_2=\frac{6^{\mathrm{o}} \times 0.5}{0.75} \\
\mathrm{~A}_2=4^{\mathrm{o}}
\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282778
The angle of a prism is \(60^{\circ}\) and the angle of minimum deviation of light passing through it is observed to be \(40^{\circ}\). The angle of incidence of light is
1 \(30^{\circ}\)
2 \(40^{\circ}\)
3 \(50^{\circ}\)
4 \(60^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}=60^{\circ}\) and \(\delta_{\mathrm{m}}=40^{\circ}\)
We know that,
Angle of incident \((i)=\frac{A+\delta_m}{2}\)
\(\begin{aligned}
i=\frac{60^{\circ}+40^{\circ}}{2} \\
i=50^{\circ}
\end{aligned}\)
AP EAMCET-1993
Ray Optics
282674
Find the value of the angle of emergence from the prism. Refractive index of the glass is \(\sqrt{3}\).
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A: Given, Refractive index of glass \((\mu)=\sqrt{3}\)
From figure -
\(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}=30^{\circ}\)
\(\left(\because \mathrm{r}_1=0^{\circ}\right)\)
\(r_2=30^{\circ}\)
From Snell's law \(\mu_1 \sin r_2=\mu_0\) sin \(\mathrm{e}\) \(\sqrt{3} \times \sin 30^{\circ}=1 \times \sin \mathrm{e}\)
\(e=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) e=60^{\circ}\)
NEET- 2021
Ray Optics
282675
A prism of refractive index \(\sqrt{2}\) has a refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers a minimum deviation?
1 \(45^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
A: Given, \(\mu=\sqrt{2}\)
Prism angle \((A)=60^{\circ}\)
For minimum deviation -
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For minimum deviation angle of incidence is equal to angle of emergence
Hence, \(i=\frac{A+\delta_m}{2}\)
Then, \(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \frac{60^{\circ}}{2}}\)
\(\sin \mathrm{i}=\sqrt{2} \times \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}\)
\(\sin \mathrm{i}=\frac{1}{\sqrt{2}}\)
\(\mathrm{i}=45^{\circ}\)
TRIPURA-2021
Ray Optics
282711
The angle of minimum deviation for a prism of angle \(A\) is \(180^{\circ}-2 A\). The refractive index is
1 \(\sin \frac{A}{2}\)
2 \(\cos \frac{\mathrm{A}}{2}\)
3 \(\tan \frac{\mathrm{A}}{2}\)
4 \(\cot \frac{\mathrm{A}}{2}\)
Explanation:
D: Given, \(\delta_{\mathrm{m}}=180^{\circ}-2 \mathrm{~A}\)
We know,
\(\begin{aligned}
\text { Refractive index }(\mu)=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)} \\
=\frac{\sin \left(\frac{\mathrm{A}+180^{\circ}-2 \mathrm{~A}}{2}\right)}{\sin \mathrm{A} / 2} \\
\mu=\cot \mathrm{A} / 2 \\
\end{aligned}\)
Manipal UGET-2017
Ray Optics
282676
A thin prism of angle \(6^{\circ}\) made up of glass of refractive index 1.5 is combined with another prism made up of glass of refractive index 1.75 to produce dispersion without deviation. Then find the angle of the second prism.
1 \(7^{\circ}\)
2 \(9^{\circ}\)
3 \(4^{\circ}\)
4 \(5^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}_1=6^{\circ}, \mu_1=1.5\)
\(\mathrm{A}_2=?, \quad \mu_2=1.75\)
We know that \(\delta=(\mu-1) \mathrm{A}\)
Given here \(\quad \delta_1=\delta_2\)
\(\begin{aligned}
\mathrm{A}_1\left(\mu_1-1\right)=\mathrm{A}_2\left(\mu_2-1\right) \\
6^{\mathrm{o}}(1.5-1)=\mathrm{A}_2(1.75-1) \\
\mathrm{A}_2=\frac{6^{\mathrm{o}} \times 0.5}{0.75} \\
\mathrm{~A}_2=4^{\mathrm{o}}
\end{aligned}\)
282778
The angle of a prism is \(60^{\circ}\) and the angle of minimum deviation of light passing through it is observed to be \(40^{\circ}\). The angle of incidence of light is
1 \(30^{\circ}\)
2 \(40^{\circ}\)
3 \(50^{\circ}\)
4 \(60^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}=60^{\circ}\) and \(\delta_{\mathrm{m}}=40^{\circ}\)
We know that,
Angle of incident \((i)=\frac{A+\delta_m}{2}\)
\(\begin{aligned}
i=\frac{60^{\circ}+40^{\circ}}{2} \\
i=50^{\circ}
\end{aligned}\)
AP EAMCET-1993
Ray Optics
282674
Find the value of the angle of emergence from the prism. Refractive index of the glass is \(\sqrt{3}\).
1 \(60^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A: Given, Refractive index of glass \((\mu)=\sqrt{3}\)
From figure -
\(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}=30^{\circ}\)
\(\left(\because \mathrm{r}_1=0^{\circ}\right)\)
\(r_2=30^{\circ}\)
From Snell's law \(\mu_1 \sin r_2=\mu_0\) sin \(\mathrm{e}\) \(\sqrt{3} \times \sin 30^{\circ}=1 \times \sin \mathrm{e}\)
\(e=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) e=60^{\circ}\)
NEET- 2021
Ray Optics
282675
A prism of refractive index \(\sqrt{2}\) has a refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers a minimum deviation?
1 \(45^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
A: Given, \(\mu=\sqrt{2}\)
Prism angle \((A)=60^{\circ}\)
For minimum deviation -
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For minimum deviation angle of incidence is equal to angle of emergence
Hence, \(i=\frac{A+\delta_m}{2}\)
Then, \(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \frac{60^{\circ}}{2}}\)
\(\sin \mathrm{i}=\sqrt{2} \times \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}\)
\(\sin \mathrm{i}=\frac{1}{\sqrt{2}}\)
\(\mathrm{i}=45^{\circ}\)
TRIPURA-2021
Ray Optics
282711
The angle of minimum deviation for a prism of angle \(A\) is \(180^{\circ}-2 A\). The refractive index is
1 \(\sin \frac{A}{2}\)
2 \(\cos \frac{\mathrm{A}}{2}\)
3 \(\tan \frac{\mathrm{A}}{2}\)
4 \(\cot \frac{\mathrm{A}}{2}\)
Explanation:
D: Given, \(\delta_{\mathrm{m}}=180^{\circ}-2 \mathrm{~A}\)
We know,
\(\begin{aligned}
\text { Refractive index }(\mu)=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)} \\
=\frac{\sin \left(\frac{\mathrm{A}+180^{\circ}-2 \mathrm{~A}}{2}\right)}{\sin \mathrm{A} / 2} \\
\mu=\cot \mathrm{A} / 2 \\
\end{aligned}\)
Manipal UGET-2017
Ray Optics
282676
A thin prism of angle \(6^{\circ}\) made up of glass of refractive index 1.5 is combined with another prism made up of glass of refractive index 1.75 to produce dispersion without deviation. Then find the angle of the second prism.
1 \(7^{\circ}\)
2 \(9^{\circ}\)
3 \(4^{\circ}\)
4 \(5^{\circ}\)
Explanation:
C: Given, \(\mathrm{A}_1=6^{\circ}, \mu_1=1.5\)
\(\mathrm{A}_2=?, \quad \mu_2=1.75\)
We know that \(\delta=(\mu-1) \mathrm{A}\)
Given here \(\quad \delta_1=\delta_2\)
\(\begin{aligned}
\mathrm{A}_1\left(\mu_1-1\right)=\mathrm{A}_2\left(\mu_2-1\right) \\
6^{\mathrm{o}}(1.5-1)=\mathrm{A}_2(1.75-1) \\
\mathrm{A}_2=\frac{6^{\mathrm{o}} \times 0.5}{0.75} \\
\mathrm{~A}_2=4^{\mathrm{o}}
\end{aligned}\)
