282927
Resolving power of the telescope will be more, if the diameter of the objective is
1 larger
2 smaller
3 it does not depends on diameter
4 None of these
Explanation:
A: Resolving power \((\alpha)=\frac{\mathrm{d}}{1.22 \lambda}\)
Where,
\(\mathrm{d}=\text { diameter of aperture }\)
Hence,
\(\alpha \propto d\)
Hence, resolving power of the telescope will be more, if the diameter of the objective is larger.
BITSAT-2014
Ray Optics
282870
The limit resolution of an oil immersion objective microscope of numerical aperture 0.8 for light of wavelength \(0.6 \mu \mathrm{m}\) is
282871
A telescope has an objective of focal length 100 \(\mathrm{cm}\) and an eye-piece of focal length \(5 \mathrm{~cm}\). The magnifying power of the telescope is
1 20
2 500
3 \(\frac{1}{20}\)
4 105
Explanation:
A: Given,
Focal length of object \(\left(f_o\right)=100 \mathrm{~cm}\)
Focal length of eye \(\left(f_e\right)=5 \mathrm{~cm}\)
Magnification \((\mathrm{m})=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}=\frac{100}{5}=20\)
TS- EAMCET-10.09.2020
Ray Optics
282872
A convex lens and a concave lens, each with focal length of \(4 \mathrm{~cm}\) are separated by a distance of \(6 \mathrm{~cm}\) along their axis. An object is placed 8 cm before the convex lens. The distance between the object and its image is
1 \(10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}\)
4 \(24 \mathrm{~cm}\)
Explanation:
C: Given,
Case-I
Focal length of convex lens \(\left(f_1\right)=4 \mathrm{~cm}\)
Object distance \(\left(u_1\right)=-8 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{4}=\frac{1}{\mathrm{v}_1}+\frac{1}{8} \\
\frac{1}{\mathrm{v}_1}=\frac{1}{4}-\frac{1}{8} \\
\mathrm{v}_1=8 \mathrm{~cm}
\end{aligned}\)
Now, this image will be act as object for concave lens.
Case-II
\(\therefore\) Focal length of concave lens \(=-4 \mathrm{~cm}\) (given)
Again,
\(\text { Object distance }\left(u_2\right)=8-6=2 \mathrm{~cm}\)
Using lens formula
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2} \\
\frac{1}{-4}=\frac{1}{\mathrm{v}_2}-\frac{1}{2} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{2}-\frac{1}{4} \\
\mathrm{v}_2=4 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Distance between object and final image will be
\(\begin{aligned}
\mathrm{d}=8+6+4 \\
\mathrm{~d}=18 \mathrm{~cm}
\end{aligned}\)
TS- EAMCET-10.09.2020
Ray Optics
282873
The limit of resolution of a telescope is \(3.0 \times 10^{-7} \mathrm{rad}\). Assuming that it is used to see the light of wavelength \(525 \mathrm{~nm}\) from a star, what should be the diameter of the objective?
282927
Resolving power of the telescope will be more, if the diameter of the objective is
1 larger
2 smaller
3 it does not depends on diameter
4 None of these
Explanation:
A: Resolving power \((\alpha)=\frac{\mathrm{d}}{1.22 \lambda}\)
Where,
\(\mathrm{d}=\text { diameter of aperture }\)
Hence,
\(\alpha \propto d\)
Hence, resolving power of the telescope will be more, if the diameter of the objective is larger.
BITSAT-2014
Ray Optics
282870
The limit resolution of an oil immersion objective microscope of numerical aperture 0.8 for light of wavelength \(0.6 \mu \mathrm{m}\) is
282871
A telescope has an objective of focal length 100 \(\mathrm{cm}\) and an eye-piece of focal length \(5 \mathrm{~cm}\). The magnifying power of the telescope is
1 20
2 500
3 \(\frac{1}{20}\)
4 105
Explanation:
A: Given,
Focal length of object \(\left(f_o\right)=100 \mathrm{~cm}\)
Focal length of eye \(\left(f_e\right)=5 \mathrm{~cm}\)
Magnification \((\mathrm{m})=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}=\frac{100}{5}=20\)
TS- EAMCET-10.09.2020
Ray Optics
282872
A convex lens and a concave lens, each with focal length of \(4 \mathrm{~cm}\) are separated by a distance of \(6 \mathrm{~cm}\) along their axis. An object is placed 8 cm before the convex lens. The distance between the object and its image is
1 \(10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}\)
4 \(24 \mathrm{~cm}\)
Explanation:
C: Given,
Case-I
Focal length of convex lens \(\left(f_1\right)=4 \mathrm{~cm}\)
Object distance \(\left(u_1\right)=-8 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{4}=\frac{1}{\mathrm{v}_1}+\frac{1}{8} \\
\frac{1}{\mathrm{v}_1}=\frac{1}{4}-\frac{1}{8} \\
\mathrm{v}_1=8 \mathrm{~cm}
\end{aligned}\)
Now, this image will be act as object for concave lens.
Case-II
\(\therefore\) Focal length of concave lens \(=-4 \mathrm{~cm}\) (given)
Again,
\(\text { Object distance }\left(u_2\right)=8-6=2 \mathrm{~cm}\)
Using lens formula
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2} \\
\frac{1}{-4}=\frac{1}{\mathrm{v}_2}-\frac{1}{2} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{2}-\frac{1}{4} \\
\mathrm{v}_2=4 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Distance between object and final image will be
\(\begin{aligned}
\mathrm{d}=8+6+4 \\
\mathrm{~d}=18 \mathrm{~cm}
\end{aligned}\)
TS- EAMCET-10.09.2020
Ray Optics
282873
The limit of resolution of a telescope is \(3.0 \times 10^{-7} \mathrm{rad}\). Assuming that it is used to see the light of wavelength \(525 \mathrm{~nm}\) from a star, what should be the diameter of the objective?
282927
Resolving power of the telescope will be more, if the diameter of the objective is
1 larger
2 smaller
3 it does not depends on diameter
4 None of these
Explanation:
A: Resolving power \((\alpha)=\frac{\mathrm{d}}{1.22 \lambda}\)
Where,
\(\mathrm{d}=\text { diameter of aperture }\)
Hence,
\(\alpha \propto d\)
Hence, resolving power of the telescope will be more, if the diameter of the objective is larger.
BITSAT-2014
Ray Optics
282870
The limit resolution of an oil immersion objective microscope of numerical aperture 0.8 for light of wavelength \(0.6 \mu \mathrm{m}\) is
282871
A telescope has an objective of focal length 100 \(\mathrm{cm}\) and an eye-piece of focal length \(5 \mathrm{~cm}\). The magnifying power of the telescope is
1 20
2 500
3 \(\frac{1}{20}\)
4 105
Explanation:
A: Given,
Focal length of object \(\left(f_o\right)=100 \mathrm{~cm}\)
Focal length of eye \(\left(f_e\right)=5 \mathrm{~cm}\)
Magnification \((\mathrm{m})=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}=\frac{100}{5}=20\)
TS- EAMCET-10.09.2020
Ray Optics
282872
A convex lens and a concave lens, each with focal length of \(4 \mathrm{~cm}\) are separated by a distance of \(6 \mathrm{~cm}\) along their axis. An object is placed 8 cm before the convex lens. The distance between the object and its image is
1 \(10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}\)
4 \(24 \mathrm{~cm}\)
Explanation:
C: Given,
Case-I
Focal length of convex lens \(\left(f_1\right)=4 \mathrm{~cm}\)
Object distance \(\left(u_1\right)=-8 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{4}=\frac{1}{\mathrm{v}_1}+\frac{1}{8} \\
\frac{1}{\mathrm{v}_1}=\frac{1}{4}-\frac{1}{8} \\
\mathrm{v}_1=8 \mathrm{~cm}
\end{aligned}\)
Now, this image will be act as object for concave lens.
Case-II
\(\therefore\) Focal length of concave lens \(=-4 \mathrm{~cm}\) (given)
Again,
\(\text { Object distance }\left(u_2\right)=8-6=2 \mathrm{~cm}\)
Using lens formula
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2} \\
\frac{1}{-4}=\frac{1}{\mathrm{v}_2}-\frac{1}{2} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{2}-\frac{1}{4} \\
\mathrm{v}_2=4 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Distance between object and final image will be
\(\begin{aligned}
\mathrm{d}=8+6+4 \\
\mathrm{~d}=18 \mathrm{~cm}
\end{aligned}\)
TS- EAMCET-10.09.2020
Ray Optics
282873
The limit of resolution of a telescope is \(3.0 \times 10^{-7} \mathrm{rad}\). Assuming that it is used to see the light of wavelength \(525 \mathrm{~nm}\) from a star, what should be the diameter of the objective?
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282927
Resolving power of the telescope will be more, if the diameter of the objective is
1 larger
2 smaller
3 it does not depends on diameter
4 None of these
Explanation:
A: Resolving power \((\alpha)=\frac{\mathrm{d}}{1.22 \lambda}\)
Where,
\(\mathrm{d}=\text { diameter of aperture }\)
Hence,
\(\alpha \propto d\)
Hence, resolving power of the telescope will be more, if the diameter of the objective is larger.
BITSAT-2014
Ray Optics
282870
The limit resolution of an oil immersion objective microscope of numerical aperture 0.8 for light of wavelength \(0.6 \mu \mathrm{m}\) is
282871
A telescope has an objective of focal length 100 \(\mathrm{cm}\) and an eye-piece of focal length \(5 \mathrm{~cm}\). The magnifying power of the telescope is
1 20
2 500
3 \(\frac{1}{20}\)
4 105
Explanation:
A: Given,
Focal length of object \(\left(f_o\right)=100 \mathrm{~cm}\)
Focal length of eye \(\left(f_e\right)=5 \mathrm{~cm}\)
Magnification \((\mathrm{m})=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}=\frac{100}{5}=20\)
TS- EAMCET-10.09.2020
Ray Optics
282872
A convex lens and a concave lens, each with focal length of \(4 \mathrm{~cm}\) are separated by a distance of \(6 \mathrm{~cm}\) along their axis. An object is placed 8 cm before the convex lens. The distance between the object and its image is
1 \(10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}\)
4 \(24 \mathrm{~cm}\)
Explanation:
C: Given,
Case-I
Focal length of convex lens \(\left(f_1\right)=4 \mathrm{~cm}\)
Object distance \(\left(u_1\right)=-8 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{4}=\frac{1}{\mathrm{v}_1}+\frac{1}{8} \\
\frac{1}{\mathrm{v}_1}=\frac{1}{4}-\frac{1}{8} \\
\mathrm{v}_1=8 \mathrm{~cm}
\end{aligned}\)
Now, this image will be act as object for concave lens.
Case-II
\(\therefore\) Focal length of concave lens \(=-4 \mathrm{~cm}\) (given)
Again,
\(\text { Object distance }\left(u_2\right)=8-6=2 \mathrm{~cm}\)
Using lens formula
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2} \\
\frac{1}{-4}=\frac{1}{\mathrm{v}_2}-\frac{1}{2} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{2}-\frac{1}{4} \\
\mathrm{v}_2=4 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Distance between object and final image will be
\(\begin{aligned}
\mathrm{d}=8+6+4 \\
\mathrm{~d}=18 \mathrm{~cm}
\end{aligned}\)
TS- EAMCET-10.09.2020
Ray Optics
282873
The limit of resolution of a telescope is \(3.0 \times 10^{-7} \mathrm{rad}\). Assuming that it is used to see the light of wavelength \(525 \mathrm{~nm}\) from a star, what should be the diameter of the objective?
282927
Resolving power of the telescope will be more, if the diameter of the objective is
1 larger
2 smaller
3 it does not depends on diameter
4 None of these
Explanation:
A: Resolving power \((\alpha)=\frac{\mathrm{d}}{1.22 \lambda}\)
Where,
\(\mathrm{d}=\text { diameter of aperture }\)
Hence,
\(\alpha \propto d\)
Hence, resolving power of the telescope will be more, if the diameter of the objective is larger.
BITSAT-2014
Ray Optics
282870
The limit resolution of an oil immersion objective microscope of numerical aperture 0.8 for light of wavelength \(0.6 \mu \mathrm{m}\) is
282871
A telescope has an objective of focal length 100 \(\mathrm{cm}\) and an eye-piece of focal length \(5 \mathrm{~cm}\). The magnifying power of the telescope is
1 20
2 500
3 \(\frac{1}{20}\)
4 105
Explanation:
A: Given,
Focal length of object \(\left(f_o\right)=100 \mathrm{~cm}\)
Focal length of eye \(\left(f_e\right)=5 \mathrm{~cm}\)
Magnification \((\mathrm{m})=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}=\frac{100}{5}=20\)
TS- EAMCET-10.09.2020
Ray Optics
282872
A convex lens and a concave lens, each with focal length of \(4 \mathrm{~cm}\) are separated by a distance of \(6 \mathrm{~cm}\) along their axis. An object is placed 8 cm before the convex lens. The distance between the object and its image is
1 \(10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}\)
4 \(24 \mathrm{~cm}\)
Explanation:
C: Given,
Case-I
Focal length of convex lens \(\left(f_1\right)=4 \mathrm{~cm}\)
Object distance \(\left(u_1\right)=-8 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}_1}=\frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}_1} \\
\frac{1}{4}=\frac{1}{\mathrm{v}_1}+\frac{1}{8} \\
\frac{1}{\mathrm{v}_1}=\frac{1}{4}-\frac{1}{8} \\
\mathrm{v}_1=8 \mathrm{~cm}
\end{aligned}\)
Now, this image will be act as object for concave lens.
Case-II
\(\therefore\) Focal length of concave lens \(=-4 \mathrm{~cm}\) (given)
Again,
\(\text { Object distance }\left(u_2\right)=8-6=2 \mathrm{~cm}\)
Using lens formula
\(\begin{aligned}
\frac{1}{\mathrm{f}_2}=\frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2} \\
\frac{1}{-4}=\frac{1}{\mathrm{v}_2}-\frac{1}{2} \\
\frac{1}{\mathrm{v}_2}=\frac{1}{2}-\frac{1}{4} \\
\mathrm{v}_2=4 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Distance between object and final image will be
\(\begin{aligned}
\mathrm{d}=8+6+4 \\
\mathrm{~d}=18 \mathrm{~cm}
\end{aligned}\)
TS- EAMCET-10.09.2020
Ray Optics
282873
The limit of resolution of a telescope is \(3.0 \times 10^{-7} \mathrm{rad}\). Assuming that it is used to see the light of wavelength \(525 \mathrm{~nm}\) from a star, what should be the diameter of the objective?