282460
A converging lens forms a real image \(I\) on its optic axis. A rectangular glass slab of refractive inde \(x \mu\) and thickness \(t\) is introduced between the lens and the image \(I\). The image I will move
1 away from the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
2 towards the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
3 away from the lens by \(t(\mu-1)\)
4 towards the lens by \(t(\mu-1)\)
Explanation:
A: Due to insertion of slab, the optical path increased by \(\frac{t}{\mu}\), where \(t\) is thickness of slab. Therefore converging point will shift away by -
\(\left(t-\frac{t}{\mu}\right)=t\left(1-\frac{1}{\mu}\right)\)
CG PET- 2017
Ray Optics
282461
A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin Plano-convex lens of focal length 15 \(\mathrm{cm}\), if the plane surface is silvered. The image will form at
1 \(60 \mathrm{~cm}\) left of \(\mathrm{AB}\)
2 \(30 \mathrm{~cm}\) left of \(\mathrm{AB}\)
3 \(12 \mathrm{~cm}\) left of \(A B\)
4 \(60 \mathrm{~cm}\) right of \(A B\)
Explanation:
C: Given that, \(\mathrm{u}=-20 \mathrm{~cm}\)
Focus length of plano convex lens is \(\frac{\mathrm{f}}{2}=\frac{-15}{2}\)
Using mirror formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
\frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{2}{-15} \\
\frac{1}{\mathrm{v}}=\frac{-2}{15}+\frac{1}{20} \\
\frac{1}{\mathrm{v}}=\frac{-8+3}{60} \\
\mathrm{v}=-12 \mathrm{~cm}
\end{aligned}\)
\(12 \mathrm{~cm}\) left of \(\mathrm{AB}\).
Manipal UGET-2014
Ray Optics
282462
The Plano-convex lens of focal length \(20 \mathrm{~cm}\) and \(30 \mathrm{~cm}\) are placed together to form a double convex lens, the final focal length will be
1 \(12 \mathrm{~cm}\)
2 \(60 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
A: Given that,
Focal length, \(\mathrm{f}_1=20 \mathrm{~cm}\)
\(\mathrm{f}_2=30 \mathrm{~cm}\)
We know that equivalent focal length,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{\mathrm{f}}=\frac{1}{20}+\frac{1}{30} \\
\frac{1}{\mathrm{f}}=\frac{5}{60} \\
\mathrm{~F}=12 \mathrm{~cm} \\
&
\end{aligned}\)
Manipal UGET-2013
Ray Optics
282463
Power of a lens is - 4D and for second lens, power is \(+2 \mathrm{D}\); the total power of the couple is
1 \(-2 \mathrm{D}\)
2 \(6 \mathrm{D}\)
3 \(-6 \mathrm{D}\)
4 \(-8 \mathrm{D}\)
Explanation:
A: Given that,
Power \(\left(P_1\right)=-4 \mathrm{D}\), Power \(\left(\mathrm{P}_2\right)=+2 \mathrm{D}\)
We know that,
Combination of power of lens
\(\begin{aligned}
P=P_1+P_2 \\
P=-4+2 \\
P=-2 D
\end{aligned}\)
282460
A converging lens forms a real image \(I\) on its optic axis. A rectangular glass slab of refractive inde \(x \mu\) and thickness \(t\) is introduced between the lens and the image \(I\). The image I will move
1 away from the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
2 towards the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
3 away from the lens by \(t(\mu-1)\)
4 towards the lens by \(t(\mu-1)\)
Explanation:
A: Due to insertion of slab, the optical path increased by \(\frac{t}{\mu}\), where \(t\) is thickness of slab. Therefore converging point will shift away by -
\(\left(t-\frac{t}{\mu}\right)=t\left(1-\frac{1}{\mu}\right)\)
CG PET- 2017
Ray Optics
282461
A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin Plano-convex lens of focal length 15 \(\mathrm{cm}\), if the plane surface is silvered. The image will form at
1 \(60 \mathrm{~cm}\) left of \(\mathrm{AB}\)
2 \(30 \mathrm{~cm}\) left of \(\mathrm{AB}\)
3 \(12 \mathrm{~cm}\) left of \(A B\)
4 \(60 \mathrm{~cm}\) right of \(A B\)
Explanation:
C: Given that, \(\mathrm{u}=-20 \mathrm{~cm}\)
Focus length of plano convex lens is \(\frac{\mathrm{f}}{2}=\frac{-15}{2}\)
Using mirror formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
\frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{2}{-15} \\
\frac{1}{\mathrm{v}}=\frac{-2}{15}+\frac{1}{20} \\
\frac{1}{\mathrm{v}}=\frac{-8+3}{60} \\
\mathrm{v}=-12 \mathrm{~cm}
\end{aligned}\)
\(12 \mathrm{~cm}\) left of \(\mathrm{AB}\).
Manipal UGET-2014
Ray Optics
282462
The Plano-convex lens of focal length \(20 \mathrm{~cm}\) and \(30 \mathrm{~cm}\) are placed together to form a double convex lens, the final focal length will be
1 \(12 \mathrm{~cm}\)
2 \(60 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
A: Given that,
Focal length, \(\mathrm{f}_1=20 \mathrm{~cm}\)
\(\mathrm{f}_2=30 \mathrm{~cm}\)
We know that equivalent focal length,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{\mathrm{f}}=\frac{1}{20}+\frac{1}{30} \\
\frac{1}{\mathrm{f}}=\frac{5}{60} \\
\mathrm{~F}=12 \mathrm{~cm} \\
&
\end{aligned}\)
Manipal UGET-2013
Ray Optics
282463
Power of a lens is - 4D and for second lens, power is \(+2 \mathrm{D}\); the total power of the couple is
1 \(-2 \mathrm{D}\)
2 \(6 \mathrm{D}\)
3 \(-6 \mathrm{D}\)
4 \(-8 \mathrm{D}\)
Explanation:
A: Given that,
Power \(\left(P_1\right)=-4 \mathrm{D}\), Power \(\left(\mathrm{P}_2\right)=+2 \mathrm{D}\)
We know that,
Combination of power of lens
\(\begin{aligned}
P=P_1+P_2 \\
P=-4+2 \\
P=-2 D
\end{aligned}\)
282460
A converging lens forms a real image \(I\) on its optic axis. A rectangular glass slab of refractive inde \(x \mu\) and thickness \(t\) is introduced between the lens and the image \(I\). The image I will move
1 away from the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
2 towards the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
3 away from the lens by \(t(\mu-1)\)
4 towards the lens by \(t(\mu-1)\)
Explanation:
A: Due to insertion of slab, the optical path increased by \(\frac{t}{\mu}\), where \(t\) is thickness of slab. Therefore converging point will shift away by -
\(\left(t-\frac{t}{\mu}\right)=t\left(1-\frac{1}{\mu}\right)\)
CG PET- 2017
Ray Optics
282461
A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin Plano-convex lens of focal length 15 \(\mathrm{cm}\), if the plane surface is silvered. The image will form at
1 \(60 \mathrm{~cm}\) left of \(\mathrm{AB}\)
2 \(30 \mathrm{~cm}\) left of \(\mathrm{AB}\)
3 \(12 \mathrm{~cm}\) left of \(A B\)
4 \(60 \mathrm{~cm}\) right of \(A B\)
Explanation:
C: Given that, \(\mathrm{u}=-20 \mathrm{~cm}\)
Focus length of plano convex lens is \(\frac{\mathrm{f}}{2}=\frac{-15}{2}\)
Using mirror formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
\frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{2}{-15} \\
\frac{1}{\mathrm{v}}=\frac{-2}{15}+\frac{1}{20} \\
\frac{1}{\mathrm{v}}=\frac{-8+3}{60} \\
\mathrm{v}=-12 \mathrm{~cm}
\end{aligned}\)
\(12 \mathrm{~cm}\) left of \(\mathrm{AB}\).
Manipal UGET-2014
Ray Optics
282462
The Plano-convex lens of focal length \(20 \mathrm{~cm}\) and \(30 \mathrm{~cm}\) are placed together to form a double convex lens, the final focal length will be
1 \(12 \mathrm{~cm}\)
2 \(60 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
A: Given that,
Focal length, \(\mathrm{f}_1=20 \mathrm{~cm}\)
\(\mathrm{f}_2=30 \mathrm{~cm}\)
We know that equivalent focal length,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{\mathrm{f}}=\frac{1}{20}+\frac{1}{30} \\
\frac{1}{\mathrm{f}}=\frac{5}{60} \\
\mathrm{~F}=12 \mathrm{~cm} \\
&
\end{aligned}\)
Manipal UGET-2013
Ray Optics
282463
Power of a lens is - 4D and for second lens, power is \(+2 \mathrm{D}\); the total power of the couple is
1 \(-2 \mathrm{D}\)
2 \(6 \mathrm{D}\)
3 \(-6 \mathrm{D}\)
4 \(-8 \mathrm{D}\)
Explanation:
A: Given that,
Power \(\left(P_1\right)=-4 \mathrm{D}\), Power \(\left(\mathrm{P}_2\right)=+2 \mathrm{D}\)
We know that,
Combination of power of lens
\(\begin{aligned}
P=P_1+P_2 \\
P=-4+2 \\
P=-2 D
\end{aligned}\)
282460
A converging lens forms a real image \(I\) on its optic axis. A rectangular glass slab of refractive inde \(x \mu\) and thickness \(t\) is introduced between the lens and the image \(I\). The image I will move
1 away from the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
2 towards the lens by \(\mathrm{t}\left[1-\frac{1}{\mu}\right]\)
3 away from the lens by \(t(\mu-1)\)
4 towards the lens by \(t(\mu-1)\)
Explanation:
A: Due to insertion of slab, the optical path increased by \(\frac{t}{\mu}\), where \(t\) is thickness of slab. Therefore converging point will shift away by -
\(\left(t-\frac{t}{\mu}\right)=t\left(1-\frac{1}{\mu}\right)\)
CG PET- 2017
Ray Optics
282461
A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin Plano-convex lens of focal length 15 \(\mathrm{cm}\), if the plane surface is silvered. The image will form at
1 \(60 \mathrm{~cm}\) left of \(\mathrm{AB}\)
2 \(30 \mathrm{~cm}\) left of \(\mathrm{AB}\)
3 \(12 \mathrm{~cm}\) left of \(A B\)
4 \(60 \mathrm{~cm}\) right of \(A B\)
Explanation:
C: Given that, \(\mathrm{u}=-20 \mathrm{~cm}\)
Focus length of plano convex lens is \(\frac{\mathrm{f}}{2}=\frac{-15}{2}\)
Using mirror formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
\frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{2}{-15} \\
\frac{1}{\mathrm{v}}=\frac{-2}{15}+\frac{1}{20} \\
\frac{1}{\mathrm{v}}=\frac{-8+3}{60} \\
\mathrm{v}=-12 \mathrm{~cm}
\end{aligned}\)
\(12 \mathrm{~cm}\) left of \(\mathrm{AB}\).
Manipal UGET-2014
Ray Optics
282462
The Plano-convex lens of focal length \(20 \mathrm{~cm}\) and \(30 \mathrm{~cm}\) are placed together to form a double convex lens, the final focal length will be
1 \(12 \mathrm{~cm}\)
2 \(60 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
A: Given that,
Focal length, \(\mathrm{f}_1=20 \mathrm{~cm}\)
\(\mathrm{f}_2=30 \mathrm{~cm}\)
We know that equivalent focal length,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{\mathrm{f}}=\frac{1}{20}+\frac{1}{30} \\
\frac{1}{\mathrm{f}}=\frac{5}{60} \\
\mathrm{~F}=12 \mathrm{~cm} \\
&
\end{aligned}\)
Manipal UGET-2013
Ray Optics
282463
Power of a lens is - 4D and for second lens, power is \(+2 \mathrm{D}\); the total power of the couple is
1 \(-2 \mathrm{D}\)
2 \(6 \mathrm{D}\)
3 \(-6 \mathrm{D}\)
4 \(-8 \mathrm{D}\)
Explanation:
A: Given that,
Power \(\left(P_1\right)=-4 \mathrm{D}\), Power \(\left(\mathrm{P}_2\right)=+2 \mathrm{D}\)
We know that,
Combination of power of lens
\(\begin{aligned}
P=P_1+P_2 \\
P=-4+2 \\
P=-2 D
\end{aligned}\)
