282396
An object is placed at a certain distance left to a convex lens of focal length \(20 \mathrm{~cm}\). Find the distance of the object if the image obtained is magnified by 4 times.
282397
The position of final image formed by the given lens combination from the third lens will be at a distance of \(\left(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\right.\) and \(f_3=\) \(+30 \mathrm{~cm}\) )
1 \(15 \mathrm{~cm}\)
2 infinity
3 \(45 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
D: Given,
\(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\) and \(f_3=+30 \mathrm{~cm}\) Using lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{30} \\
\mathrm{v}_1=15 \mathrm{~cm} \quad \text { (behind the lens) }
\end{aligned}\)
(behind the lens)
For second lens-
\(\begin{aligned}
\mathrm{u}_2=15-5=10 \mathrm{~cm} \\
\text { and } \mathrm{f}_2=-10 \mathrm{~cm}
\end{aligned}\)
The ray will merge parallel to axis as the virtual object is at focus of concave lens, as shown. Image of lens first will be at infinity. The parallel rays are incident on third lens of convex lens, \(\mathrm{f}_3=30 \mathrm{~cm}\).
\(\therefore\) Image formed by third lens is at focus, \(\mathrm{f}_3=30 \mathrm{~cm}\).
TS- EAMCET-09.09.2020
Ray Optics
282398
The magnifications produced by a convex lens for two position of an object are 4 and 3, respectively. If the distance of separation between the two positions of the object is \(2 \mathrm{~cm}\), then the focal length of the lens is
282400
How can we change a camera from \(\mathrm{F} / 4\) to \(\mathrm{F} / 5.6\) ?
1 Increase the aperture to 2 time keeping the focal distance constant.
2 Increase the aperture to \(\sqrt{2}\) time keeping the focal distance constant.
3 Increase the aperture to \(\frac{1}{2}\) time keeping the focal distance constant.
4 Increase the aperture to \(\frac{1}{\sqrt{2}}\) time keeping the focal distance constant.
Explanation:
D: For F/4 camera,
\(\mathrm{F}-\text { number of camera }=\frac{\text { Focal length of lens }}{\text { diameter of aperture }}\)
For \(\mathrm{F} / 4\) camera \(\mathrm{F}\) number is 4 ,
\(\therefore \quad 4=\frac{\mathrm{f}}{\mathrm{D}_1}\)
Similarly for \(\mathrm{F} / 5.6\) camera, \(\mathrm{F}\) number is 5.6
\(\therefore 5.6=\frac{\mathrm{f}}{\mathrm{D}_2}\)
Dividing equation (i) by (ii),
\(\begin{aligned}
\frac{D_2}{D_1}=\frac{4}{5.6}=\frac{1}{\sqrt{2}} \\
D_2=\frac{D_1}{\sqrt{2}}
\end{aligned}\)
282396
An object is placed at a certain distance left to a convex lens of focal length \(20 \mathrm{~cm}\). Find the distance of the object if the image obtained is magnified by 4 times.
282397
The position of final image formed by the given lens combination from the third lens will be at a distance of \(\left(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\right.\) and \(f_3=\) \(+30 \mathrm{~cm}\) )
1 \(15 \mathrm{~cm}\)
2 infinity
3 \(45 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
D: Given,
\(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\) and \(f_3=+30 \mathrm{~cm}\) Using lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{30} \\
\mathrm{v}_1=15 \mathrm{~cm} \quad \text { (behind the lens) }
\end{aligned}\)
(behind the lens)
For second lens-
\(\begin{aligned}
\mathrm{u}_2=15-5=10 \mathrm{~cm} \\
\text { and } \mathrm{f}_2=-10 \mathrm{~cm}
\end{aligned}\)
The ray will merge parallel to axis as the virtual object is at focus of concave lens, as shown. Image of lens first will be at infinity. The parallel rays are incident on third lens of convex lens, \(\mathrm{f}_3=30 \mathrm{~cm}\).
\(\therefore\) Image formed by third lens is at focus, \(\mathrm{f}_3=30 \mathrm{~cm}\).
TS- EAMCET-09.09.2020
Ray Optics
282398
The magnifications produced by a convex lens for two position of an object are 4 and 3, respectively. If the distance of separation between the two positions of the object is \(2 \mathrm{~cm}\), then the focal length of the lens is
282400
How can we change a camera from \(\mathrm{F} / 4\) to \(\mathrm{F} / 5.6\) ?
1 Increase the aperture to 2 time keeping the focal distance constant.
2 Increase the aperture to \(\sqrt{2}\) time keeping the focal distance constant.
3 Increase the aperture to \(\frac{1}{2}\) time keeping the focal distance constant.
4 Increase the aperture to \(\frac{1}{\sqrt{2}}\) time keeping the focal distance constant.
Explanation:
D: For F/4 camera,
\(\mathrm{F}-\text { number of camera }=\frac{\text { Focal length of lens }}{\text { diameter of aperture }}\)
For \(\mathrm{F} / 4\) camera \(\mathrm{F}\) number is 4 ,
\(\therefore \quad 4=\frac{\mathrm{f}}{\mathrm{D}_1}\)
Similarly for \(\mathrm{F} / 5.6\) camera, \(\mathrm{F}\) number is 5.6
\(\therefore 5.6=\frac{\mathrm{f}}{\mathrm{D}_2}\)
Dividing equation (i) by (ii),
\(\begin{aligned}
\frac{D_2}{D_1}=\frac{4}{5.6}=\frac{1}{\sqrt{2}} \\
D_2=\frac{D_1}{\sqrt{2}}
\end{aligned}\)
282396
An object is placed at a certain distance left to a convex lens of focal length \(20 \mathrm{~cm}\). Find the distance of the object if the image obtained is magnified by 4 times.
282397
The position of final image formed by the given lens combination from the third lens will be at a distance of \(\left(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\right.\) and \(f_3=\) \(+30 \mathrm{~cm}\) )
1 \(15 \mathrm{~cm}\)
2 infinity
3 \(45 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
D: Given,
\(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\) and \(f_3=+30 \mathrm{~cm}\) Using lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{30} \\
\mathrm{v}_1=15 \mathrm{~cm} \quad \text { (behind the lens) }
\end{aligned}\)
(behind the lens)
For second lens-
\(\begin{aligned}
\mathrm{u}_2=15-5=10 \mathrm{~cm} \\
\text { and } \mathrm{f}_2=-10 \mathrm{~cm}
\end{aligned}\)
The ray will merge parallel to axis as the virtual object is at focus of concave lens, as shown. Image of lens first will be at infinity. The parallel rays are incident on third lens of convex lens, \(\mathrm{f}_3=30 \mathrm{~cm}\).
\(\therefore\) Image formed by third lens is at focus, \(\mathrm{f}_3=30 \mathrm{~cm}\).
TS- EAMCET-09.09.2020
Ray Optics
282398
The magnifications produced by a convex lens for two position of an object are 4 and 3, respectively. If the distance of separation between the two positions of the object is \(2 \mathrm{~cm}\), then the focal length of the lens is
282400
How can we change a camera from \(\mathrm{F} / 4\) to \(\mathrm{F} / 5.6\) ?
1 Increase the aperture to 2 time keeping the focal distance constant.
2 Increase the aperture to \(\sqrt{2}\) time keeping the focal distance constant.
3 Increase the aperture to \(\frac{1}{2}\) time keeping the focal distance constant.
4 Increase the aperture to \(\frac{1}{\sqrt{2}}\) time keeping the focal distance constant.
Explanation:
D: For F/4 camera,
\(\mathrm{F}-\text { number of camera }=\frac{\text { Focal length of lens }}{\text { diameter of aperture }}\)
For \(\mathrm{F} / 4\) camera \(\mathrm{F}\) number is 4 ,
\(\therefore \quad 4=\frac{\mathrm{f}}{\mathrm{D}_1}\)
Similarly for \(\mathrm{F} / 5.6\) camera, \(\mathrm{F}\) number is 5.6
\(\therefore 5.6=\frac{\mathrm{f}}{\mathrm{D}_2}\)
Dividing equation (i) by (ii),
\(\begin{aligned}
\frac{D_2}{D_1}=\frac{4}{5.6}=\frac{1}{\sqrt{2}} \\
D_2=\frac{D_1}{\sqrt{2}}
\end{aligned}\)
282396
An object is placed at a certain distance left to a convex lens of focal length \(20 \mathrm{~cm}\). Find the distance of the object if the image obtained is magnified by 4 times.
282397
The position of final image formed by the given lens combination from the third lens will be at a distance of \(\left(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\right.\) and \(f_3=\) \(+30 \mathrm{~cm}\) )
1 \(15 \mathrm{~cm}\)
2 infinity
3 \(45 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\)
Explanation:
D: Given,
\(f_1=+10 \mathrm{~cm}, f_2=-10 \mathrm{~cm}\) and \(f_3=+30 \mathrm{~cm}\) Using lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{10}=\frac{1}{\mathrm{v}_1}+\frac{1}{30} \\
\mathrm{v}_1=15 \mathrm{~cm} \quad \text { (behind the lens) }
\end{aligned}\)
(behind the lens)
For second lens-
\(\begin{aligned}
\mathrm{u}_2=15-5=10 \mathrm{~cm} \\
\text { and } \mathrm{f}_2=-10 \mathrm{~cm}
\end{aligned}\)
The ray will merge parallel to axis as the virtual object is at focus of concave lens, as shown. Image of lens first will be at infinity. The parallel rays are incident on third lens of convex lens, \(\mathrm{f}_3=30 \mathrm{~cm}\).
\(\therefore\) Image formed by third lens is at focus, \(\mathrm{f}_3=30 \mathrm{~cm}\).
TS- EAMCET-09.09.2020
Ray Optics
282398
The magnifications produced by a convex lens for two position of an object are 4 and 3, respectively. If the distance of separation between the two positions of the object is \(2 \mathrm{~cm}\), then the focal length of the lens is
282400
How can we change a camera from \(\mathrm{F} / 4\) to \(\mathrm{F} / 5.6\) ?
1 Increase the aperture to 2 time keeping the focal distance constant.
2 Increase the aperture to \(\sqrt{2}\) time keeping the focal distance constant.
3 Increase the aperture to \(\frac{1}{2}\) time keeping the focal distance constant.
4 Increase the aperture to \(\frac{1}{\sqrt{2}}\) time keeping the focal distance constant.
Explanation:
D: For F/4 camera,
\(\mathrm{F}-\text { number of camera }=\frac{\text { Focal length of lens }}{\text { diameter of aperture }}\)
For \(\mathrm{F} / 4\) camera \(\mathrm{F}\) number is 4 ,
\(\therefore \quad 4=\frac{\mathrm{f}}{\mathrm{D}_1}\)
Similarly for \(\mathrm{F} / 5.6\) camera, \(\mathrm{F}\) number is 5.6
\(\therefore 5.6=\frac{\mathrm{f}}{\mathrm{D}_2}\)
Dividing equation (i) by (ii),
\(\begin{aligned}
\frac{D_2}{D_1}=\frac{4}{5.6}=\frac{1}{\sqrt{2}} \\
D_2=\frac{D_1}{\sqrt{2}}
\end{aligned}\)
