282406
Find the position of final of image from first lens. Given focal length of each lens is \(\mathbf{1 0} \mathrm{cm}\).
1 \(40 \mathrm{~cm}\)
2 \(50 \mathrm{~cm}\)
3 \(45 \mathrm{~cm}\)
4 \(55 \mathrm{~cm}\)
Explanation:
B: Given, focal length of each lens is \(10 \mathrm{~cm}\) Using lens formula,
For lens I-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{40} \quad\left\{\mathrm{u}_1=-40 \mathrm{~cm} \text { given }\right\}
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}=\frac{1}{10}-\frac{1}{40} \\
\frac{1}{\mathrm{v}_1}=\frac{4-1}{40} \\
\mathrm{v}_1=\frac{40}{3} \mathrm{~cm}
\end{aligned}\)
For lens II -
\(\mathrm{f}_2=10 \mathrm{~cm}, \mathrm{u}_1=-\left(30-\frac{40}{3}\right) \mathrm{cm}\)
Again, using lens formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{10}-\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{5-3}{50} \\
\mathrm{v}_2=25 \mathrm{~cm}
\end{aligned}\)
For lens III -
\(\mathrm{f}_3=10 \mathrm{~cm}, \mathrm{u}_3=-(30-25)=-5 \mathrm{~cm}\)
Again, Using lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_3}=\frac{1}{\mathrm{v}_3}-\frac{1}{\mathrm{u}_3} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_3}+\frac{1}{5} \\
\frac{1}{\mathrm{v}_3}=\frac{1}{10}-\frac{1}{5} \\
\mathrm{v}_3=-10 \mathrm{~cm}
\end{aligned}\)
Now, distance of final image from \(\mathrm{I}^{\text {st }}\) lens is-
\(\begin{array}{ll}
\mathrm{d}=30+30-10 & \text { \{from fig.\} } \\
\mathrm{d}=50 \mathrm{~cm} &
\end{array}\)
\{from fig.\}
JIPMER-2019
Ray Optics
282407
The distance between the object and the real image formed by a convex lens is \(d\). If the magnification is \(m\), the focal length of the lens in terms of \(d\) and \(m\).
1 \(\frac{\mathrm{md}}{(\mathrm{m}+1)^2}\)
2 \(\frac{\mathrm{md}}{(\mathrm{m}+1)}\)
3 \(\frac{\mathrm{md}}{(\mathrm{m}-1)^2}\)
4 \(\frac{\mathrm{md}}{(\mathrm{m}-1)}\)
Explanation:
A: Given, The distance between object and image \((u+v)=d\) ...(i)
And, we know \(\frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\)
\(\mathrm{v}=\mathrm{um}\)
From equation (i)-
\(\begin{aligned}
\mathrm{u}+\mathrm{um}=\mathrm{d} \\
\mathrm{u}(1+\mathrm{m})=\mathrm{d} \\
\mathrm{u}=\frac{\mathrm{d}}{1+\mathrm{m}}
\end{aligned}\)
Again from equation (i),
\(\begin{aligned}
\mathrm{v}=\mathrm{d}-\mathrm{u} \\
\mathrm{v}=\mathrm{d}-\frac{\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{d}(1+\mathrm{m})-\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{dm}}{1+\mathrm{m}}
\end{aligned}\)
From lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{(-\mathrm{u})} \text { (from sign convention) }
\end{aligned}\)
Putting the value of equation (ii) and (iii) in equation (iv), we get-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1+\mathrm{m}}{\mathrm{dm}}+\frac{1+\mathrm{m}}{\mathrm{d}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~d}}\left(\frac{1+\mathrm{m}+\mathrm{m}+\mathrm{m}^2}{\mathrm{~m}}\right) \\
\frac{1}{\mathrm{f}}=\frac{(1+\mathrm{m})^2}{\mathrm{dm}} \\
\mathrm{f}=\frac{\mathrm{md}}{(\mathrm{m}+1)^2}
\end{aligned}\)
Assam CEE-2019
Ray Optics
282408
An object is seen through a simple microscope of focal length \(10 \mathrm{~cm}\). Find the angular magnification produced, if the image is formed at the near point of the eye which is \(20 \mathrm{~cm}\) away from it.
282409
Two thin lenses are in contact and the focal length of the combination is \(80 \mathrm{~cm}\). If the focal length of one lens is \(20 \mathrm{~cm}\), then the power of the other lens will be
282410
A convex lens of focal length ' \(f\) ' is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination is
282406
Find the position of final of image from first lens. Given focal length of each lens is \(\mathbf{1 0} \mathrm{cm}\).
1 \(40 \mathrm{~cm}\)
2 \(50 \mathrm{~cm}\)
3 \(45 \mathrm{~cm}\)
4 \(55 \mathrm{~cm}\)
Explanation:
B: Given, focal length of each lens is \(10 \mathrm{~cm}\) Using lens formula,
For lens I-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{40} \quad\left\{\mathrm{u}_1=-40 \mathrm{~cm} \text { given }\right\}
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}=\frac{1}{10}-\frac{1}{40} \\
\frac{1}{\mathrm{v}_1}=\frac{4-1}{40} \\
\mathrm{v}_1=\frac{40}{3} \mathrm{~cm}
\end{aligned}\)
For lens II -
\(\mathrm{f}_2=10 \mathrm{~cm}, \mathrm{u}_1=-\left(30-\frac{40}{3}\right) \mathrm{cm}\)
Again, using lens formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{10}-\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{5-3}{50} \\
\mathrm{v}_2=25 \mathrm{~cm}
\end{aligned}\)
For lens III -
\(\mathrm{f}_3=10 \mathrm{~cm}, \mathrm{u}_3=-(30-25)=-5 \mathrm{~cm}\)
Again, Using lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_3}=\frac{1}{\mathrm{v}_3}-\frac{1}{\mathrm{u}_3} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_3}+\frac{1}{5} \\
\frac{1}{\mathrm{v}_3}=\frac{1}{10}-\frac{1}{5} \\
\mathrm{v}_3=-10 \mathrm{~cm}
\end{aligned}\)
Now, distance of final image from \(\mathrm{I}^{\text {st }}\) lens is-
\(\begin{array}{ll}
\mathrm{d}=30+30-10 & \text { \{from fig.\} } \\
\mathrm{d}=50 \mathrm{~cm} &
\end{array}\)
\{from fig.\}
JIPMER-2019
Ray Optics
282407
The distance between the object and the real image formed by a convex lens is \(d\). If the magnification is \(m\), the focal length of the lens in terms of \(d\) and \(m\).
1 \(\frac{\mathrm{md}}{(\mathrm{m}+1)^2}\)
2 \(\frac{\mathrm{md}}{(\mathrm{m}+1)}\)
3 \(\frac{\mathrm{md}}{(\mathrm{m}-1)^2}\)
4 \(\frac{\mathrm{md}}{(\mathrm{m}-1)}\)
Explanation:
A: Given, The distance between object and image \((u+v)=d\) ...(i)
And, we know \(\frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\)
\(\mathrm{v}=\mathrm{um}\)
From equation (i)-
\(\begin{aligned}
\mathrm{u}+\mathrm{um}=\mathrm{d} \\
\mathrm{u}(1+\mathrm{m})=\mathrm{d} \\
\mathrm{u}=\frac{\mathrm{d}}{1+\mathrm{m}}
\end{aligned}\)
Again from equation (i),
\(\begin{aligned}
\mathrm{v}=\mathrm{d}-\mathrm{u} \\
\mathrm{v}=\mathrm{d}-\frac{\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{d}(1+\mathrm{m})-\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{dm}}{1+\mathrm{m}}
\end{aligned}\)
From lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{(-\mathrm{u})} \text { (from sign convention) }
\end{aligned}\)
Putting the value of equation (ii) and (iii) in equation (iv), we get-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1+\mathrm{m}}{\mathrm{dm}}+\frac{1+\mathrm{m}}{\mathrm{d}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~d}}\left(\frac{1+\mathrm{m}+\mathrm{m}+\mathrm{m}^2}{\mathrm{~m}}\right) \\
\frac{1}{\mathrm{f}}=\frac{(1+\mathrm{m})^2}{\mathrm{dm}} \\
\mathrm{f}=\frac{\mathrm{md}}{(\mathrm{m}+1)^2}
\end{aligned}\)
Assam CEE-2019
Ray Optics
282408
An object is seen through a simple microscope of focal length \(10 \mathrm{~cm}\). Find the angular magnification produced, if the image is formed at the near point of the eye which is \(20 \mathrm{~cm}\) away from it.
282409
Two thin lenses are in contact and the focal length of the combination is \(80 \mathrm{~cm}\). If the focal length of one lens is \(20 \mathrm{~cm}\), then the power of the other lens will be
282410
A convex lens of focal length ' \(f\) ' is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282406
Find the position of final of image from first lens. Given focal length of each lens is \(\mathbf{1 0} \mathrm{cm}\).
1 \(40 \mathrm{~cm}\)
2 \(50 \mathrm{~cm}\)
3 \(45 \mathrm{~cm}\)
4 \(55 \mathrm{~cm}\)
Explanation:
B: Given, focal length of each lens is \(10 \mathrm{~cm}\) Using lens formula,
For lens I-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{40} \quad\left\{\mathrm{u}_1=-40 \mathrm{~cm} \text { given }\right\}
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}=\frac{1}{10}-\frac{1}{40} \\
\frac{1}{\mathrm{v}_1}=\frac{4-1}{40} \\
\mathrm{v}_1=\frac{40}{3} \mathrm{~cm}
\end{aligned}\)
For lens II -
\(\mathrm{f}_2=10 \mathrm{~cm}, \mathrm{u}_1=-\left(30-\frac{40}{3}\right) \mathrm{cm}\)
Again, using lens formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{10}-\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{5-3}{50} \\
\mathrm{v}_2=25 \mathrm{~cm}
\end{aligned}\)
For lens III -
\(\mathrm{f}_3=10 \mathrm{~cm}, \mathrm{u}_3=-(30-25)=-5 \mathrm{~cm}\)
Again, Using lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_3}=\frac{1}{\mathrm{v}_3}-\frac{1}{\mathrm{u}_3} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_3}+\frac{1}{5} \\
\frac{1}{\mathrm{v}_3}=\frac{1}{10}-\frac{1}{5} \\
\mathrm{v}_3=-10 \mathrm{~cm}
\end{aligned}\)
Now, distance of final image from \(\mathrm{I}^{\text {st }}\) lens is-
\(\begin{array}{ll}
\mathrm{d}=30+30-10 & \text { \{from fig.\} } \\
\mathrm{d}=50 \mathrm{~cm} &
\end{array}\)
\{from fig.\}
JIPMER-2019
Ray Optics
282407
The distance between the object and the real image formed by a convex lens is \(d\). If the magnification is \(m\), the focal length of the lens in terms of \(d\) and \(m\).
1 \(\frac{\mathrm{md}}{(\mathrm{m}+1)^2}\)
2 \(\frac{\mathrm{md}}{(\mathrm{m}+1)}\)
3 \(\frac{\mathrm{md}}{(\mathrm{m}-1)^2}\)
4 \(\frac{\mathrm{md}}{(\mathrm{m}-1)}\)
Explanation:
A: Given, The distance between object and image \((u+v)=d\) ...(i)
And, we know \(\frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\)
\(\mathrm{v}=\mathrm{um}\)
From equation (i)-
\(\begin{aligned}
\mathrm{u}+\mathrm{um}=\mathrm{d} \\
\mathrm{u}(1+\mathrm{m})=\mathrm{d} \\
\mathrm{u}=\frac{\mathrm{d}}{1+\mathrm{m}}
\end{aligned}\)
Again from equation (i),
\(\begin{aligned}
\mathrm{v}=\mathrm{d}-\mathrm{u} \\
\mathrm{v}=\mathrm{d}-\frac{\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{d}(1+\mathrm{m})-\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{dm}}{1+\mathrm{m}}
\end{aligned}\)
From lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{(-\mathrm{u})} \text { (from sign convention) }
\end{aligned}\)
Putting the value of equation (ii) and (iii) in equation (iv), we get-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1+\mathrm{m}}{\mathrm{dm}}+\frac{1+\mathrm{m}}{\mathrm{d}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~d}}\left(\frac{1+\mathrm{m}+\mathrm{m}+\mathrm{m}^2}{\mathrm{~m}}\right) \\
\frac{1}{\mathrm{f}}=\frac{(1+\mathrm{m})^2}{\mathrm{dm}} \\
\mathrm{f}=\frac{\mathrm{md}}{(\mathrm{m}+1)^2}
\end{aligned}\)
Assam CEE-2019
Ray Optics
282408
An object is seen through a simple microscope of focal length \(10 \mathrm{~cm}\). Find the angular magnification produced, if the image is formed at the near point of the eye which is \(20 \mathrm{~cm}\) away from it.
282409
Two thin lenses are in contact and the focal length of the combination is \(80 \mathrm{~cm}\). If the focal length of one lens is \(20 \mathrm{~cm}\), then the power of the other lens will be
282410
A convex lens of focal length ' \(f\) ' is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination is
282406
Find the position of final of image from first lens. Given focal length of each lens is \(\mathbf{1 0} \mathrm{cm}\).
1 \(40 \mathrm{~cm}\)
2 \(50 \mathrm{~cm}\)
3 \(45 \mathrm{~cm}\)
4 \(55 \mathrm{~cm}\)
Explanation:
B: Given, focal length of each lens is \(10 \mathrm{~cm}\) Using lens formula,
For lens I-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{40} \quad\left\{\mathrm{u}_1=-40 \mathrm{~cm} \text { given }\right\}
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}=\frac{1}{10}-\frac{1}{40} \\
\frac{1}{\mathrm{v}_1}=\frac{4-1}{40} \\
\mathrm{v}_1=\frac{40}{3} \mathrm{~cm}
\end{aligned}\)
For lens II -
\(\mathrm{f}_2=10 \mathrm{~cm}, \mathrm{u}_1=-\left(30-\frac{40}{3}\right) \mathrm{cm}\)
Again, using lens formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{10}-\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{5-3}{50} \\
\mathrm{v}_2=25 \mathrm{~cm}
\end{aligned}\)
For lens III -
\(\mathrm{f}_3=10 \mathrm{~cm}, \mathrm{u}_3=-(30-25)=-5 \mathrm{~cm}\)
Again, Using lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_3}=\frac{1}{\mathrm{v}_3}-\frac{1}{\mathrm{u}_3} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_3}+\frac{1}{5} \\
\frac{1}{\mathrm{v}_3}=\frac{1}{10}-\frac{1}{5} \\
\mathrm{v}_3=-10 \mathrm{~cm}
\end{aligned}\)
Now, distance of final image from \(\mathrm{I}^{\text {st }}\) lens is-
\(\begin{array}{ll}
\mathrm{d}=30+30-10 & \text { \{from fig.\} } \\
\mathrm{d}=50 \mathrm{~cm} &
\end{array}\)
\{from fig.\}
JIPMER-2019
Ray Optics
282407
The distance between the object and the real image formed by a convex lens is \(d\). If the magnification is \(m\), the focal length of the lens in terms of \(d\) and \(m\).
1 \(\frac{\mathrm{md}}{(\mathrm{m}+1)^2}\)
2 \(\frac{\mathrm{md}}{(\mathrm{m}+1)}\)
3 \(\frac{\mathrm{md}}{(\mathrm{m}-1)^2}\)
4 \(\frac{\mathrm{md}}{(\mathrm{m}-1)}\)
Explanation:
A: Given, The distance between object and image \((u+v)=d\) ...(i)
And, we know \(\frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\)
\(\mathrm{v}=\mathrm{um}\)
From equation (i)-
\(\begin{aligned}
\mathrm{u}+\mathrm{um}=\mathrm{d} \\
\mathrm{u}(1+\mathrm{m})=\mathrm{d} \\
\mathrm{u}=\frac{\mathrm{d}}{1+\mathrm{m}}
\end{aligned}\)
Again from equation (i),
\(\begin{aligned}
\mathrm{v}=\mathrm{d}-\mathrm{u} \\
\mathrm{v}=\mathrm{d}-\frac{\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{d}(1+\mathrm{m})-\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{dm}}{1+\mathrm{m}}
\end{aligned}\)
From lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{(-\mathrm{u})} \text { (from sign convention) }
\end{aligned}\)
Putting the value of equation (ii) and (iii) in equation (iv), we get-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1+\mathrm{m}}{\mathrm{dm}}+\frac{1+\mathrm{m}}{\mathrm{d}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~d}}\left(\frac{1+\mathrm{m}+\mathrm{m}+\mathrm{m}^2}{\mathrm{~m}}\right) \\
\frac{1}{\mathrm{f}}=\frac{(1+\mathrm{m})^2}{\mathrm{dm}} \\
\mathrm{f}=\frac{\mathrm{md}}{(\mathrm{m}+1)^2}
\end{aligned}\)
Assam CEE-2019
Ray Optics
282408
An object is seen through a simple microscope of focal length \(10 \mathrm{~cm}\). Find the angular magnification produced, if the image is formed at the near point of the eye which is \(20 \mathrm{~cm}\) away from it.
282409
Two thin lenses are in contact and the focal length of the combination is \(80 \mathrm{~cm}\). If the focal length of one lens is \(20 \mathrm{~cm}\), then the power of the other lens will be
282410
A convex lens of focal length ' \(f\) ' is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination is
282406
Find the position of final of image from first lens. Given focal length of each lens is \(\mathbf{1 0} \mathrm{cm}\).
1 \(40 \mathrm{~cm}\)
2 \(50 \mathrm{~cm}\)
3 \(45 \mathrm{~cm}\)
4 \(55 \mathrm{~cm}\)
Explanation:
B: Given, focal length of each lens is \(10 \mathrm{~cm}\) Using lens formula,
For lens I-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{40} \quad\left\{\mathrm{u}_1=-40 \mathrm{~cm} \text { given }\right\}
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{v}_1}=\frac{1}{10}-\frac{1}{40} \\
\frac{1}{\mathrm{v}_1}=\frac{4-1}{40} \\
\mathrm{v}_1=\frac{40}{3} \mathrm{~cm}
\end{aligned}\)
For lens II -
\(\mathrm{f}_2=10 \mathrm{~cm}, \mathrm{u}_1=-\left(30-\frac{40}{3}\right) \mathrm{cm}\)
Again, using lens formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_2}+\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{10}-\frac{3}{50} \\
\frac{1}{\mathrm{v}_2}=\frac{5-3}{50} \\
\mathrm{v}_2=25 \mathrm{~cm}
\end{aligned}\)
For lens III -
\(\mathrm{f}_3=10 \mathrm{~cm}, \mathrm{u}_3=-(30-25)=-5 \mathrm{~cm}\)
Again, Using lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}_3}=\frac{1}{\mathrm{v}_3}-\frac{1}{\mathrm{u}_3} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_3}+\frac{1}{5} \\
\frac{1}{\mathrm{v}_3}=\frac{1}{10}-\frac{1}{5} \\
\mathrm{v}_3=-10 \mathrm{~cm}
\end{aligned}\)
Now, distance of final image from \(\mathrm{I}^{\text {st }}\) lens is-
\(\begin{array}{ll}
\mathrm{d}=30+30-10 & \text { \{from fig.\} } \\
\mathrm{d}=50 \mathrm{~cm} &
\end{array}\)
\{from fig.\}
JIPMER-2019
Ray Optics
282407
The distance between the object and the real image formed by a convex lens is \(d\). If the magnification is \(m\), the focal length of the lens in terms of \(d\) and \(m\).
1 \(\frac{\mathrm{md}}{(\mathrm{m}+1)^2}\)
2 \(\frac{\mathrm{md}}{(\mathrm{m}+1)}\)
3 \(\frac{\mathrm{md}}{(\mathrm{m}-1)^2}\)
4 \(\frac{\mathrm{md}}{(\mathrm{m}-1)}\)
Explanation:
A: Given, The distance between object and image \((u+v)=d\) ...(i)
And, we know \(\frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\)
\(\mathrm{v}=\mathrm{um}\)
From equation (i)-
\(\begin{aligned}
\mathrm{u}+\mathrm{um}=\mathrm{d} \\
\mathrm{u}(1+\mathrm{m})=\mathrm{d} \\
\mathrm{u}=\frac{\mathrm{d}}{1+\mathrm{m}}
\end{aligned}\)
Again from equation (i),
\(\begin{aligned}
\mathrm{v}=\mathrm{d}-\mathrm{u} \\
\mathrm{v}=\mathrm{d}-\frac{\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{d}(1+\mathrm{m})-\mathrm{d}}{1+\mathrm{m}} \\
\mathrm{v}=\frac{\mathrm{dm}}{1+\mathrm{m}}
\end{aligned}\)
From lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{(-\mathrm{u})} \text { (from sign convention) }
\end{aligned}\)
Putting the value of equation (ii) and (iii) in equation (iv), we get-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1+\mathrm{m}}{\mathrm{dm}}+\frac{1+\mathrm{m}}{\mathrm{d}} \\
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~d}}\left(\frac{1+\mathrm{m}+\mathrm{m}+\mathrm{m}^2}{\mathrm{~m}}\right) \\
\frac{1}{\mathrm{f}}=\frac{(1+\mathrm{m})^2}{\mathrm{dm}} \\
\mathrm{f}=\frac{\mathrm{md}}{(\mathrm{m}+1)^2}
\end{aligned}\)
Assam CEE-2019
Ray Optics
282408
An object is seen through a simple microscope of focal length \(10 \mathrm{~cm}\). Find the angular magnification produced, if the image is formed at the near point of the eye which is \(20 \mathrm{~cm}\) away from it.
282409
Two thin lenses are in contact and the focal length of the combination is \(80 \mathrm{~cm}\). If the focal length of one lens is \(20 \mathrm{~cm}\), then the power of the other lens will be
282410
A convex lens of focal length ' \(f\) ' is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination is
