282293
A transparent cube of \(15 \mathrm{~cm}\) edge contains a small air bubble, Its apparent depth when viewed through one face is \(6 \mathrm{~cm}\) and when viewed through the opposite face is \(4 \mathrm{~cm}\). Then the refractive index of the material of the cube is
1 2.0
2 2.5
3 1.6
4 1.5
Explanation:
D: Given,
Edge \((\mathrm{a})=15 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_1\right)=6 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_2\right)=4 \mathrm{~cm}\)
Refractive index \((\mu)=\frac{\text { Real depth }(x)}{\text { Apparant depth }\left(y_1\right)}\)
\(\mu=\frac{x}{6}\)
\(\begin{aligned}
\mu=\frac{15-x}{y_2} \\
\mu=\frac{15-x}{4}
\end{aligned}\)
Dividing equation, (i) by (ii)
\(\begin{aligned}
\frac{x}{6}=\frac{15-x}{4} \\
4 x=90-6 x \\
x=9 \mathrm{~cm}
\end{aligned}\)
From equation (i)
\(\begin{aligned}
\mu=\frac{9}{6} \\
\mu=1.5
\end{aligned}\)
UP CPMT-2004
Ray Optics
282294
A thin equiconvex lens of refractive index \(\frac{3}{2}\) and radius of curvature \(30 \mathrm{~cm}\) is put in water \(\left(\right.\) refractive index \(\left.=\frac{4}{3}\right)\), its focal length is
1 \(0.15 \mathrm{~m}\)
2 \(0.30 \mathrm{~m}\)
3 \(0.45 \mathrm{~m}\)
4 \(1.20 \mathrm{~m}\)
Explanation:
D: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=\frac{3}{2}\),
Radius of curvature \(\left(R_1\right)=30 \mathrm{~cm}\) and \(\left(R_2\right)=-30 \mathrm{~cm}\) (equiconvex lens)
Refractive index of water \(\left(\mu_w\right)=\frac{4}{3}\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{30}-\frac{1}{(-30)}\right) \\
\frac{1}{\mathrm{f}}=\frac{1}{8} \times \frac{1}{15} \\
\mathrm{f}=120 \mathrm{~cm}=1.20 \mathrm{~m}
\end{aligned}\)
UP CPMT-2012
Ray Optics
282295
The deviation is maximum for which colour
1 violet
2 red
3 blue
4 green
Explanation:
A: We know that,
Angle of deviation \((\delta) \propto \frac{1}{\text { Wavelength }}\)
And least wavelength is of violet light, hence it deflects maximum.
UP CPMT-2005
Ray Optics
282296
When a ray of light enters from one medium to another, its velocity is doubled. The critical angle for the ray for total internal reflection will be
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 information is incomplete
Explanation:
A:
\begin{tabular}{l} Given, \(\mathrm{v}_1=\mathrm{v}_2 \times 2\) \\
\(\mathrm{v}_1=\) velocity in medium 1. \\
\(\mathrm{v}_2=\) velocity in medium 2 . \\
Refractive index \((\mu)=\frac{\mathrm{v}_1}{\mathrm{v}_2}\) \\
\(\mu=\frac{2 \mathrm{v}_2}{\mathrm{v}_2}=2\) \\
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\) \\
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\frac{1}{2}\right)\)
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\sin 30^{\circ}\right)\)
\(\theta_{\mathrm{c}} =30^{\circ}\)
\end{tabular}
282293
A transparent cube of \(15 \mathrm{~cm}\) edge contains a small air bubble, Its apparent depth when viewed through one face is \(6 \mathrm{~cm}\) and when viewed through the opposite face is \(4 \mathrm{~cm}\). Then the refractive index of the material of the cube is
1 2.0
2 2.5
3 1.6
4 1.5
Explanation:
D: Given,
Edge \((\mathrm{a})=15 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_1\right)=6 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_2\right)=4 \mathrm{~cm}\)
Refractive index \((\mu)=\frac{\text { Real depth }(x)}{\text { Apparant depth }\left(y_1\right)}\)
\(\mu=\frac{x}{6}\)
\(\begin{aligned}
\mu=\frac{15-x}{y_2} \\
\mu=\frac{15-x}{4}
\end{aligned}\)
Dividing equation, (i) by (ii)
\(\begin{aligned}
\frac{x}{6}=\frac{15-x}{4} \\
4 x=90-6 x \\
x=9 \mathrm{~cm}
\end{aligned}\)
From equation (i)
\(\begin{aligned}
\mu=\frac{9}{6} \\
\mu=1.5
\end{aligned}\)
UP CPMT-2004
Ray Optics
282294
A thin equiconvex lens of refractive index \(\frac{3}{2}\) and radius of curvature \(30 \mathrm{~cm}\) is put in water \(\left(\right.\) refractive index \(\left.=\frac{4}{3}\right)\), its focal length is
1 \(0.15 \mathrm{~m}\)
2 \(0.30 \mathrm{~m}\)
3 \(0.45 \mathrm{~m}\)
4 \(1.20 \mathrm{~m}\)
Explanation:
D: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=\frac{3}{2}\),
Radius of curvature \(\left(R_1\right)=30 \mathrm{~cm}\) and \(\left(R_2\right)=-30 \mathrm{~cm}\) (equiconvex lens)
Refractive index of water \(\left(\mu_w\right)=\frac{4}{3}\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{30}-\frac{1}{(-30)}\right) \\
\frac{1}{\mathrm{f}}=\frac{1}{8} \times \frac{1}{15} \\
\mathrm{f}=120 \mathrm{~cm}=1.20 \mathrm{~m}
\end{aligned}\)
UP CPMT-2012
Ray Optics
282295
The deviation is maximum for which colour
1 violet
2 red
3 blue
4 green
Explanation:
A: We know that,
Angle of deviation \((\delta) \propto \frac{1}{\text { Wavelength }}\)
And least wavelength is of violet light, hence it deflects maximum.
UP CPMT-2005
Ray Optics
282296
When a ray of light enters from one medium to another, its velocity is doubled. The critical angle for the ray for total internal reflection will be
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 information is incomplete
Explanation:
A:
\begin{tabular}{l} Given, \(\mathrm{v}_1=\mathrm{v}_2 \times 2\) \\
\(\mathrm{v}_1=\) velocity in medium 1. \\
\(\mathrm{v}_2=\) velocity in medium 2 . \\
Refractive index \((\mu)=\frac{\mathrm{v}_1}{\mathrm{v}_2}\) \\
\(\mu=\frac{2 \mathrm{v}_2}{\mathrm{v}_2}=2\) \\
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\) \\
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\frac{1}{2}\right)\)
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\sin 30^{\circ}\right)\)
\(\theta_{\mathrm{c}} =30^{\circ}\)
\end{tabular}
282293
A transparent cube of \(15 \mathrm{~cm}\) edge contains a small air bubble, Its apparent depth when viewed through one face is \(6 \mathrm{~cm}\) and when viewed through the opposite face is \(4 \mathrm{~cm}\). Then the refractive index of the material of the cube is
1 2.0
2 2.5
3 1.6
4 1.5
Explanation:
D: Given,
Edge \((\mathrm{a})=15 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_1\right)=6 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_2\right)=4 \mathrm{~cm}\)
Refractive index \((\mu)=\frac{\text { Real depth }(x)}{\text { Apparant depth }\left(y_1\right)}\)
\(\mu=\frac{x}{6}\)
\(\begin{aligned}
\mu=\frac{15-x}{y_2} \\
\mu=\frac{15-x}{4}
\end{aligned}\)
Dividing equation, (i) by (ii)
\(\begin{aligned}
\frac{x}{6}=\frac{15-x}{4} \\
4 x=90-6 x \\
x=9 \mathrm{~cm}
\end{aligned}\)
From equation (i)
\(\begin{aligned}
\mu=\frac{9}{6} \\
\mu=1.5
\end{aligned}\)
UP CPMT-2004
Ray Optics
282294
A thin equiconvex lens of refractive index \(\frac{3}{2}\) and radius of curvature \(30 \mathrm{~cm}\) is put in water \(\left(\right.\) refractive index \(\left.=\frac{4}{3}\right)\), its focal length is
1 \(0.15 \mathrm{~m}\)
2 \(0.30 \mathrm{~m}\)
3 \(0.45 \mathrm{~m}\)
4 \(1.20 \mathrm{~m}\)
Explanation:
D: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=\frac{3}{2}\),
Radius of curvature \(\left(R_1\right)=30 \mathrm{~cm}\) and \(\left(R_2\right)=-30 \mathrm{~cm}\) (equiconvex lens)
Refractive index of water \(\left(\mu_w\right)=\frac{4}{3}\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{30}-\frac{1}{(-30)}\right) \\
\frac{1}{\mathrm{f}}=\frac{1}{8} \times \frac{1}{15} \\
\mathrm{f}=120 \mathrm{~cm}=1.20 \mathrm{~m}
\end{aligned}\)
UP CPMT-2012
Ray Optics
282295
The deviation is maximum for which colour
1 violet
2 red
3 blue
4 green
Explanation:
A: We know that,
Angle of deviation \((\delta) \propto \frac{1}{\text { Wavelength }}\)
And least wavelength is of violet light, hence it deflects maximum.
UP CPMT-2005
Ray Optics
282296
When a ray of light enters from one medium to another, its velocity is doubled. The critical angle for the ray for total internal reflection will be
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 information is incomplete
Explanation:
A:
\begin{tabular}{l} Given, \(\mathrm{v}_1=\mathrm{v}_2 \times 2\) \\
\(\mathrm{v}_1=\) velocity in medium 1. \\
\(\mathrm{v}_2=\) velocity in medium 2 . \\
Refractive index \((\mu)=\frac{\mathrm{v}_1}{\mathrm{v}_2}\) \\
\(\mu=\frac{2 \mathrm{v}_2}{\mathrm{v}_2}=2\) \\
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\) \\
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\frac{1}{2}\right)\)
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\sin 30^{\circ}\right)\)
\(\theta_{\mathrm{c}} =30^{\circ}\)
\end{tabular}
282293
A transparent cube of \(15 \mathrm{~cm}\) edge contains a small air bubble, Its apparent depth when viewed through one face is \(6 \mathrm{~cm}\) and when viewed through the opposite face is \(4 \mathrm{~cm}\). Then the refractive index of the material of the cube is
1 2.0
2 2.5
3 1.6
4 1.5
Explanation:
D: Given,
Edge \((\mathrm{a})=15 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_1\right)=6 \mathrm{~cm}\)
Apparent depth \(\left(\mathrm{y}_2\right)=4 \mathrm{~cm}\)
Refractive index \((\mu)=\frac{\text { Real depth }(x)}{\text { Apparant depth }\left(y_1\right)}\)
\(\mu=\frac{x}{6}\)
\(\begin{aligned}
\mu=\frac{15-x}{y_2} \\
\mu=\frac{15-x}{4}
\end{aligned}\)
Dividing equation, (i) by (ii)
\(\begin{aligned}
\frac{x}{6}=\frac{15-x}{4} \\
4 x=90-6 x \\
x=9 \mathrm{~cm}
\end{aligned}\)
From equation (i)
\(\begin{aligned}
\mu=\frac{9}{6} \\
\mu=1.5
\end{aligned}\)
UP CPMT-2004
Ray Optics
282294
A thin equiconvex lens of refractive index \(\frac{3}{2}\) and radius of curvature \(30 \mathrm{~cm}\) is put in water \(\left(\right.\) refractive index \(\left.=\frac{4}{3}\right)\), its focal length is
1 \(0.15 \mathrm{~m}\)
2 \(0.30 \mathrm{~m}\)
3 \(0.45 \mathrm{~m}\)
4 \(1.20 \mathrm{~m}\)
Explanation:
D: Given that,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=\frac{3}{2}\),
Radius of curvature \(\left(R_1\right)=30 \mathrm{~cm}\) and \(\left(R_2\right)=-30 \mathrm{~cm}\) (equiconvex lens)
Refractive index of water \(\left(\mu_w\right)=\frac{4}{3}\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{30}-\frac{1}{(-30)}\right) \\
\frac{1}{\mathrm{f}}=\frac{1}{8} \times \frac{1}{15} \\
\mathrm{f}=120 \mathrm{~cm}=1.20 \mathrm{~m}
\end{aligned}\)
UP CPMT-2012
Ray Optics
282295
The deviation is maximum for which colour
1 violet
2 red
3 blue
4 green
Explanation:
A: We know that,
Angle of deviation \((\delta) \propto \frac{1}{\text { Wavelength }}\)
And least wavelength is of violet light, hence it deflects maximum.
UP CPMT-2005
Ray Optics
282296
When a ray of light enters from one medium to another, its velocity is doubled. The critical angle for the ray for total internal reflection will be
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 information is incomplete
Explanation:
A:
\begin{tabular}{l} Given, \(\mathrm{v}_1=\mathrm{v}_2 \times 2\) \\
\(\mathrm{v}_1=\) velocity in medium 1. \\
\(\mathrm{v}_2=\) velocity in medium 2 . \\
Refractive index \((\mu)=\frac{\mathrm{v}_1}{\mathrm{v}_2}\) \\
\(\mu=\frac{2 \mathrm{v}_2}{\mathrm{v}_2}=2\) \\
Critical angle \(\left(\theta_{\mathrm{c}}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)\) \\
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\frac{1}{2}\right)\)
\(\theta_{\mathrm{c}} =\sin ^{-1}\left(\sin 30^{\circ}\right)\)
\(\theta_{\mathrm{c}} =30^{\circ}\)
\end{tabular}
