NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282686
When a ray of light is incident at an angle of \(30^{\circ}\) on an equilateral glass prism, it suffers minimum deviation. The angle between the emergent ray and the second refracting surface of the prism is
1 \(60^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(0^{\circ}\)
Explanation:
A: Given, \(\mathrm{i}=30^{\circ}\)
Under minimum deviation condition, the angle between emergent ray and second refracting surface of prism is = 2 i
\(\begin{aligned}
=2 \times 30^{\circ} \\
=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282687
When a light ray is incident on a prism at an angle of \(45^{\circ}\), the minimum deviation is obtained. If refractive index of material of prism is \(\sqrt{2}\), then angle of prism will be
\(\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\)
1 \(75^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
D: Given, \(\mathrm{i}=45^{\circ}, \mu=\sqrt{2}\)
We know, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)}\)
But, \(\quad \frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}=\mathrm{i}\)
So,
\(\begin{aligned}
\mu=\frac{\sin i}{\sin (A / 2)} \\
\sqrt{2}=\frac{\sin 45^{\circ}}{\sin (A / 2)} \\
\sin \frac{A}{2}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}
\end{aligned}\)
\(\mathrm{A}=2 \cdot \sin ^{-1}\left(\frac{1}{2}\right) \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right)\)
\(\begin{aligned}
\mathrm{A}=2 \times 30^{\circ} \\
\mathrm{A}=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282688
A ray of light suffers minimum deviation in equilateral prism \(P\). Additional prisms \(Q\) and \(R\) of identical shape and of same material as that of \(P\) are now combined as shown in figure. The ray will now suffer
1 Greater deviation
2 No deviation
3 Same deviation as before
4 Total internal reflection
Explanation:
C: There will be no refraction from P to Q and then from Q to \(\mathrm{R}\) (all being identical). Hence the ray will have same deviation as before.
COMEDK 2020
Ray Optics
282689
A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and the latter is equal to \(\frac{3}{4}\) the angle of prism. The angle of deviation is
1 \(25^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(35^{\circ}\)
Explanation:
B: Given, angle of incident \(=\) angle of emergent
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \mathrm{~A}\)
Angle of prism, \(\mathrm{A}=60^{\circ}\)
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \times 60^{\circ}=45^{\circ}\)
From prism formula,
\(\begin{aligned}
\mathrm{A}+\delta=\mathrm{i}+\mathrm{e} \\
\delta=45^{\circ}+45^{\circ}-60^{\circ} \\
\delta=90^{\circ}-60^{\circ} \\
\delta=30^{\circ}
\end{aligned}\)
So, the angle of deviation \((\delta)=30^{\circ}\)
282686
When a ray of light is incident at an angle of \(30^{\circ}\) on an equilateral glass prism, it suffers minimum deviation. The angle between the emergent ray and the second refracting surface of the prism is
1 \(60^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(0^{\circ}\)
Explanation:
A: Given, \(\mathrm{i}=30^{\circ}\)
Under minimum deviation condition, the angle between emergent ray and second refracting surface of prism is = 2 i
\(\begin{aligned}
=2 \times 30^{\circ} \\
=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282687
When a light ray is incident on a prism at an angle of \(45^{\circ}\), the minimum deviation is obtained. If refractive index of material of prism is \(\sqrt{2}\), then angle of prism will be
\(\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\)
1 \(75^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
D: Given, \(\mathrm{i}=45^{\circ}, \mu=\sqrt{2}\)
We know, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)}\)
But, \(\quad \frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}=\mathrm{i}\)
So,
\(\begin{aligned}
\mu=\frac{\sin i}{\sin (A / 2)} \\
\sqrt{2}=\frac{\sin 45^{\circ}}{\sin (A / 2)} \\
\sin \frac{A}{2}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}
\end{aligned}\)
\(\mathrm{A}=2 \cdot \sin ^{-1}\left(\frac{1}{2}\right) \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right)\)
\(\begin{aligned}
\mathrm{A}=2 \times 30^{\circ} \\
\mathrm{A}=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282688
A ray of light suffers minimum deviation in equilateral prism \(P\). Additional prisms \(Q\) and \(R\) of identical shape and of same material as that of \(P\) are now combined as shown in figure. The ray will now suffer
1 Greater deviation
2 No deviation
3 Same deviation as before
4 Total internal reflection
Explanation:
C: There will be no refraction from P to Q and then from Q to \(\mathrm{R}\) (all being identical). Hence the ray will have same deviation as before.
COMEDK 2020
Ray Optics
282689
A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and the latter is equal to \(\frac{3}{4}\) the angle of prism. The angle of deviation is
1 \(25^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(35^{\circ}\)
Explanation:
B: Given, angle of incident \(=\) angle of emergent
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \mathrm{~A}\)
Angle of prism, \(\mathrm{A}=60^{\circ}\)
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \times 60^{\circ}=45^{\circ}\)
From prism formula,
\(\begin{aligned}
\mathrm{A}+\delta=\mathrm{i}+\mathrm{e} \\
\delta=45^{\circ}+45^{\circ}-60^{\circ} \\
\delta=90^{\circ}-60^{\circ} \\
\delta=30^{\circ}
\end{aligned}\)
So, the angle of deviation \((\delta)=30^{\circ}\)
282686
When a ray of light is incident at an angle of \(30^{\circ}\) on an equilateral glass prism, it suffers minimum deviation. The angle between the emergent ray and the second refracting surface of the prism is
1 \(60^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(0^{\circ}\)
Explanation:
A: Given, \(\mathrm{i}=30^{\circ}\)
Under minimum deviation condition, the angle between emergent ray and second refracting surface of prism is = 2 i
\(\begin{aligned}
=2 \times 30^{\circ} \\
=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282687
When a light ray is incident on a prism at an angle of \(45^{\circ}\), the minimum deviation is obtained. If refractive index of material of prism is \(\sqrt{2}\), then angle of prism will be
\(\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\)
1 \(75^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
D: Given, \(\mathrm{i}=45^{\circ}, \mu=\sqrt{2}\)
We know, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)}\)
But, \(\quad \frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}=\mathrm{i}\)
So,
\(\begin{aligned}
\mu=\frac{\sin i}{\sin (A / 2)} \\
\sqrt{2}=\frac{\sin 45^{\circ}}{\sin (A / 2)} \\
\sin \frac{A}{2}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}
\end{aligned}\)
\(\mathrm{A}=2 \cdot \sin ^{-1}\left(\frac{1}{2}\right) \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right)\)
\(\begin{aligned}
\mathrm{A}=2 \times 30^{\circ} \\
\mathrm{A}=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282688
A ray of light suffers minimum deviation in equilateral prism \(P\). Additional prisms \(Q\) and \(R\) of identical shape and of same material as that of \(P\) are now combined as shown in figure. The ray will now suffer
1 Greater deviation
2 No deviation
3 Same deviation as before
4 Total internal reflection
Explanation:
C: There will be no refraction from P to Q and then from Q to \(\mathrm{R}\) (all being identical). Hence the ray will have same deviation as before.
COMEDK 2020
Ray Optics
282689
A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and the latter is equal to \(\frac{3}{4}\) the angle of prism. The angle of deviation is
1 \(25^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(35^{\circ}\)
Explanation:
B: Given, angle of incident \(=\) angle of emergent
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \mathrm{~A}\)
Angle of prism, \(\mathrm{A}=60^{\circ}\)
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \times 60^{\circ}=45^{\circ}\)
From prism formula,
\(\begin{aligned}
\mathrm{A}+\delta=\mathrm{i}+\mathrm{e} \\
\delta=45^{\circ}+45^{\circ}-60^{\circ} \\
\delta=90^{\circ}-60^{\circ} \\
\delta=30^{\circ}
\end{aligned}\)
So, the angle of deviation \((\delta)=30^{\circ}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ray Optics
282686
When a ray of light is incident at an angle of \(30^{\circ}\) on an equilateral glass prism, it suffers minimum deviation. The angle between the emergent ray and the second refracting surface of the prism is
1 \(60^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(0^{\circ}\)
Explanation:
A: Given, \(\mathrm{i}=30^{\circ}\)
Under minimum deviation condition, the angle between emergent ray and second refracting surface of prism is = 2 i
\(\begin{aligned}
=2 \times 30^{\circ} \\
=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282687
When a light ray is incident on a prism at an angle of \(45^{\circ}\), the minimum deviation is obtained. If refractive index of material of prism is \(\sqrt{2}\), then angle of prism will be
\(\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\)
1 \(75^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
D: Given, \(\mathrm{i}=45^{\circ}, \mu=\sqrt{2}\)
We know, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A / 2)}\)
But, \(\quad \frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}=\mathrm{i}\)
So,
\(\begin{aligned}
\mu=\frac{\sin i}{\sin (A / 2)} \\
\sqrt{2}=\frac{\sin 45^{\circ}}{\sin (A / 2)} \\
\sin \frac{A}{2}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}
\end{aligned}\)
\(\mathrm{A}=2 \cdot \sin ^{-1}\left(\frac{1}{2}\right) \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right)\)
\(\begin{aligned}
\mathrm{A}=2 \times 30^{\circ} \\
\mathrm{A}=60^{\circ}
\end{aligned}\)
MHT-CET 2020
Ray Optics
282688
A ray of light suffers minimum deviation in equilateral prism \(P\). Additional prisms \(Q\) and \(R\) of identical shape and of same material as that of \(P\) are now combined as shown in figure. The ray will now suffer
1 Greater deviation
2 No deviation
3 Same deviation as before
4 Total internal reflection
Explanation:
C: There will be no refraction from P to Q and then from Q to \(\mathrm{R}\) (all being identical). Hence the ray will have same deviation as before.
COMEDK 2020
Ray Optics
282689
A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and the latter is equal to \(\frac{3}{4}\) the angle of prism. The angle of deviation is
1 \(25^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(35^{\circ}\)
Explanation:
B: Given, angle of incident \(=\) angle of emergent
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \mathrm{~A}\)
Angle of prism, \(\mathrm{A}=60^{\circ}\)
\(\mathrm{i}=\mathrm{e}=\frac{3}{4} \times 60^{\circ}=45^{\circ}\)
From prism formula,
\(\begin{aligned}
\mathrm{A}+\delta=\mathrm{i}+\mathrm{e} \\
\delta=45^{\circ}+45^{\circ}-60^{\circ} \\
\delta=90^{\circ}-60^{\circ} \\
\delta=30^{\circ}
\end{aligned}\)
So, the angle of deviation \((\delta)=30^{\circ}\)
