282523
Magnitude of the focal length of lens B, in cm, is
1 30
2 25
3 20
4 10
Explanation:
A: Case-I,
Distance of image \(\left(\mathrm{v}_1\right)=20 \mathrm{~cm}\)
Distance of object \(\left(\mathrm{u}_1\right)=\infty\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\frac{1}{20}-\frac{1}{-\infty}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\mathrm{f}_{\mathrm{A}}=20 \mathrm{~cm} \\
\text { Lens B is midway }
\end{aligned}\)
Case-II
Lens B is midway
So,
\(\mathrm{u}_2=10 \mathrm{~cm}\)
\(\& \mathrm{v}_2=15 \mathrm{~cm}\) as screen is shifted by \(5 \mathrm{~cm}\) away Again,
\(\begin{aligned}
\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{1}{15}-\frac{1}{10}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{2-3}{30}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\mathrm{f}_{\mathrm{B}}=-30 \mathrm{~cm} \text { or } 30 \mathrm{~cm}
\end{aligned}\)
COMEDK 2011
Ray Optics
282525
Two convex lenses \(A\) and \(B\) placed in contact from the image of a distant object at \(P\). If the lens \(B\) is moved to the right a little, the image will
1 Move to the left
2 Move to the right
3 Remain at \(\mathrm{P}\)
4 Move either to the left or right, depending upon focal length of the lenses
Explanation:
B:
Equivalent focal length
\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{\mathrm{A}}}+\frac{1}{\mathrm{f}_{\mathrm{B}}}\)
Now, If the lens B shifted to the right
Equivalent focal length
\(\frac{1}{F^{\prime}}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{d}{f_A f_B}\)
From equation (i) and (ii)
It is clear that the \(\mathrm{F}^{\prime}<\mathrm{F}\),
So, image will shift toward right.
COMEDK 2015
Ray Optics
282526
An object is placed at a distance of \(1.5 \mathrm{~m}\) from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is
1 \(1 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.24 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
C: Given,
\(|\mathrm{u}|+|\mathrm{v}|=1.5\)
Magnification, \(\mathrm{m}=4\)
\(\begin{aligned}
\mathrm{m}=\frac{-\mathrm{V}}{\mathrm{u}} \\
4=\frac{-\mathrm{V}}{\mathrm{u}} \\
\mathrm{u}=\frac{-\mathrm{V}}{4}
\end{aligned}\)
Putting the value of \(u\) in equation (i)
\(\begin{aligned}
\frac{\mathrm{v}}{4}+\mathrm{v}=1.5 \\
\frac{5 \mathrm{v}}{4}=1.5 \\
\mathrm{v}=1.2 \mathrm{~m}
\end{aligned}\)
hence, \(\quad \mathrm{u}=\frac{-1.2}{4}=-0.3 \mathrm{~m}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{f}=\frac{1}{V}-\frac{1}{u}=\frac{1}{1.2}+\frac{1}{0.3} \\
\mathrm{f}=0.24 \mathrm{~m}
\end{aligned}\)
282523
Magnitude of the focal length of lens B, in cm, is
1 30
2 25
3 20
4 10
Explanation:
A: Case-I,
Distance of image \(\left(\mathrm{v}_1\right)=20 \mathrm{~cm}\)
Distance of object \(\left(\mathrm{u}_1\right)=\infty\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\frac{1}{20}-\frac{1}{-\infty}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\mathrm{f}_{\mathrm{A}}=20 \mathrm{~cm} \\
\text { Lens B is midway }
\end{aligned}\)
Case-II
Lens B is midway
So,
\(\mathrm{u}_2=10 \mathrm{~cm}\)
\(\& \mathrm{v}_2=15 \mathrm{~cm}\) as screen is shifted by \(5 \mathrm{~cm}\) away Again,
\(\begin{aligned}
\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{1}{15}-\frac{1}{10}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{2-3}{30}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\mathrm{f}_{\mathrm{B}}=-30 \mathrm{~cm} \text { or } 30 \mathrm{~cm}
\end{aligned}\)
COMEDK 2011
Ray Optics
282525
Two convex lenses \(A\) and \(B\) placed in contact from the image of a distant object at \(P\). If the lens \(B\) is moved to the right a little, the image will
1 Move to the left
2 Move to the right
3 Remain at \(\mathrm{P}\)
4 Move either to the left or right, depending upon focal length of the lenses
Explanation:
B:
Equivalent focal length
\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{\mathrm{A}}}+\frac{1}{\mathrm{f}_{\mathrm{B}}}\)
Now, If the lens B shifted to the right
Equivalent focal length
\(\frac{1}{F^{\prime}}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{d}{f_A f_B}\)
From equation (i) and (ii)
It is clear that the \(\mathrm{F}^{\prime}<\mathrm{F}\),
So, image will shift toward right.
COMEDK 2015
Ray Optics
282526
An object is placed at a distance of \(1.5 \mathrm{~m}\) from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is
1 \(1 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.24 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
C: Given,
\(|\mathrm{u}|+|\mathrm{v}|=1.5\)
Magnification, \(\mathrm{m}=4\)
\(\begin{aligned}
\mathrm{m}=\frac{-\mathrm{V}}{\mathrm{u}} \\
4=\frac{-\mathrm{V}}{\mathrm{u}} \\
\mathrm{u}=\frac{-\mathrm{V}}{4}
\end{aligned}\)
Putting the value of \(u\) in equation (i)
\(\begin{aligned}
\frac{\mathrm{v}}{4}+\mathrm{v}=1.5 \\
\frac{5 \mathrm{v}}{4}=1.5 \\
\mathrm{v}=1.2 \mathrm{~m}
\end{aligned}\)
hence, \(\quad \mathrm{u}=\frac{-1.2}{4}=-0.3 \mathrm{~m}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{f}=\frac{1}{V}-\frac{1}{u}=\frac{1}{1.2}+\frac{1}{0.3} \\
\mathrm{f}=0.24 \mathrm{~m}
\end{aligned}\)
282523
Magnitude of the focal length of lens B, in cm, is
1 30
2 25
3 20
4 10
Explanation:
A: Case-I,
Distance of image \(\left(\mathrm{v}_1\right)=20 \mathrm{~cm}\)
Distance of object \(\left(\mathrm{u}_1\right)=\infty\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\frac{1}{20}-\frac{1}{-\infty}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\mathrm{f}_{\mathrm{A}}=20 \mathrm{~cm} \\
\text { Lens B is midway }
\end{aligned}\)
Case-II
Lens B is midway
So,
\(\mathrm{u}_2=10 \mathrm{~cm}\)
\(\& \mathrm{v}_2=15 \mathrm{~cm}\) as screen is shifted by \(5 \mathrm{~cm}\) away Again,
\(\begin{aligned}
\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{1}{15}-\frac{1}{10}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{2-3}{30}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\mathrm{f}_{\mathrm{B}}=-30 \mathrm{~cm} \text { or } 30 \mathrm{~cm}
\end{aligned}\)
COMEDK 2011
Ray Optics
282525
Two convex lenses \(A\) and \(B\) placed in contact from the image of a distant object at \(P\). If the lens \(B\) is moved to the right a little, the image will
1 Move to the left
2 Move to the right
3 Remain at \(\mathrm{P}\)
4 Move either to the left or right, depending upon focal length of the lenses
Explanation:
B:
Equivalent focal length
\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{\mathrm{A}}}+\frac{1}{\mathrm{f}_{\mathrm{B}}}\)
Now, If the lens B shifted to the right
Equivalent focal length
\(\frac{1}{F^{\prime}}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{d}{f_A f_B}\)
From equation (i) and (ii)
It is clear that the \(\mathrm{F}^{\prime}<\mathrm{F}\),
So, image will shift toward right.
COMEDK 2015
Ray Optics
282526
An object is placed at a distance of \(1.5 \mathrm{~m}\) from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is
1 \(1 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.24 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
C: Given,
\(|\mathrm{u}|+|\mathrm{v}|=1.5\)
Magnification, \(\mathrm{m}=4\)
\(\begin{aligned}
\mathrm{m}=\frac{-\mathrm{V}}{\mathrm{u}} \\
4=\frac{-\mathrm{V}}{\mathrm{u}} \\
\mathrm{u}=\frac{-\mathrm{V}}{4}
\end{aligned}\)
Putting the value of \(u\) in equation (i)
\(\begin{aligned}
\frac{\mathrm{v}}{4}+\mathrm{v}=1.5 \\
\frac{5 \mathrm{v}}{4}=1.5 \\
\mathrm{v}=1.2 \mathrm{~m}
\end{aligned}\)
hence, \(\quad \mathrm{u}=\frac{-1.2}{4}=-0.3 \mathrm{~m}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{f}=\frac{1}{V}-\frac{1}{u}=\frac{1}{1.2}+\frac{1}{0.3} \\
\mathrm{f}=0.24 \mathrm{~m}
\end{aligned}\)
282523
Magnitude of the focal length of lens B, in cm, is
1 30
2 25
3 20
4 10
Explanation:
A: Case-I,
Distance of image \(\left(\mathrm{v}_1\right)=20 \mathrm{~cm}\)
Distance of object \(\left(\mathrm{u}_1\right)=\infty\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\frac{1}{20}-\frac{1}{-\infty}=\frac{1}{\mathrm{f}_{\mathrm{A}}} \\
\mathrm{f}_{\mathrm{A}}=20 \mathrm{~cm} \\
\text { Lens B is midway }
\end{aligned}\)
Case-II
Lens B is midway
So,
\(\mathrm{u}_2=10 \mathrm{~cm}\)
\(\& \mathrm{v}_2=15 \mathrm{~cm}\) as screen is shifted by \(5 \mathrm{~cm}\) away Again,
\(\begin{aligned}
\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{1}{15}-\frac{1}{10}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\frac{2-3}{30}=\frac{1}{\mathrm{f}_{\mathrm{B}}} \\
\mathrm{f}_{\mathrm{B}}=-30 \mathrm{~cm} \text { or } 30 \mathrm{~cm}
\end{aligned}\)
COMEDK 2011
Ray Optics
282525
Two convex lenses \(A\) and \(B\) placed in contact from the image of a distant object at \(P\). If the lens \(B\) is moved to the right a little, the image will
1 Move to the left
2 Move to the right
3 Remain at \(\mathrm{P}\)
4 Move either to the left or right, depending upon focal length of the lenses
Explanation:
B:
Equivalent focal length
\(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{\mathrm{A}}}+\frac{1}{\mathrm{f}_{\mathrm{B}}}\)
Now, If the lens B shifted to the right
Equivalent focal length
\(\frac{1}{F^{\prime}}=\frac{1}{f_A}+\frac{1}{f_B}-\frac{d}{f_A f_B}\)
From equation (i) and (ii)
It is clear that the \(\mathrm{F}^{\prime}<\mathrm{F}\),
So, image will shift toward right.
COMEDK 2015
Ray Optics
282526
An object is placed at a distance of \(1.5 \mathrm{~m}\) from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is
1 \(1 \mathrm{~m}\)
2 \(0.5 \mathrm{~m}\)
3 \(0.24 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
C: Given,
\(|\mathrm{u}|+|\mathrm{v}|=1.5\)
Magnification, \(\mathrm{m}=4\)
\(\begin{aligned}
\mathrm{m}=\frac{-\mathrm{V}}{\mathrm{u}} \\
4=\frac{-\mathrm{V}}{\mathrm{u}} \\
\mathrm{u}=\frac{-\mathrm{V}}{4}
\end{aligned}\)
Putting the value of \(u\) in equation (i)
\(\begin{aligned}
\frac{\mathrm{v}}{4}+\mathrm{v}=1.5 \\
\frac{5 \mathrm{v}}{4}=1.5 \\
\mathrm{v}=1.2 \mathrm{~m}
\end{aligned}\)
hence, \(\quad \mathrm{u}=\frac{-1.2}{4}=-0.3 \mathrm{~m}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{f}=\frac{1}{V}-\frac{1}{u}=\frac{1}{1.2}+\frac{1}{0.3} \\
\mathrm{f}=0.24 \mathrm{~m}
\end{aligned}\)
