282874
The magnifying power of a telescope is \(\mathrm{m}^{\prime}\). If the focal length of the eye piece is doubled, then its magnifying power will become
1 \(\sqrt{2} \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{\mathrm{m}}{2}\)
Explanation:
D: For a telescope, Magnifying power
\(m=\frac{\text { focal length of objective }\left(f_0\right)}{\text { focallength of eyepiece }\left(f_c\right)}\)
\(\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_e}\)
If, \(\quad \mathrm{f}_{\mathrm{s}}^{\prime}=2 \mathrm{f}_{\mathrm{c}}\)
New magnifying power,
\(\mathrm{m}^{\prime}=\frac{\mathrm{f}_o}{\mathrm{f}_e^{\prime}}=\frac{\mathrm{f}_{\mathrm{o}}}{2 \mathrm{f}_e}=\frac{\mathrm{m}}{2}\)
Manipal UGET -2020
Ray Optics
282875
Find the angular magnification of telescope if focal length of objective and eye lenses are 10 \(\mathrm{cm}\) and \(10 \mathrm{~mm}\) respectively and tube length is \(11 \mathrm{~cm}\) :
1 10
2 5
3 100
4 50
Explanation:
A: Given,
Focal length of object \(\left(f_{\mathrm{o}}\right)=10 \mathrm{~cm}\) and focal length of eye lens \(\left(\mathrm{f}_{\mathrm{c}}\right)=10 \mathrm{~mm}=1 \mathrm{~cm}\) As we know that,
Angular magnification of telescope \(=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_e}=\frac{10}{\mathrm{l}}=10\)
AIIMS-26.05.2019(E) Shift-2
Ray Optics
282876
The limit of resolution of a telescope is \(2.5 \times 10^{-}\) rad. If the telescope is used to detect light of wavelength \(500 \mathrm{~mm}\) coming from a star, the diameter of the objective lens used by telescope is
1 \(244 \mathrm{~cm}\)
2 \(258 \mathrm{~cm}\)
3 \(228 \mathrm{~cm}\)
4 \(264 \mathrm{~cm}\)
Explanation:
A: Given,
Limit of resolution \((\alpha)=2.5 \times 10^{-7}\) rad
Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\)
We know that,
Diameter of lens \((d)=\frac{1.22 \lambda}{\alpha}\)
\(\begin{aligned}
d=\frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-7}} \\
d=244 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-04.05.2019
Ray Optics
282877
The focal length of the objective lens of a telescope is \(50 \mathrm{~cm}\). If the magnification of the telescope is 25 , then the focal length of the eyepiece is
1 \(12.5 \mathrm{~cm}\)
2 \(5 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
C: Given,
Focal length of objective \(\left(f_0\right)=50 \mathrm{~cm}\)
Magnification \((\mathrm{m})=25\)
We know that.
Magnification of telescope,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}} \\
25=\frac{50}{\mathrm{f}_{\mathrm{e}}} \\
\mathrm{f}_{\mathrm{c}}=\frac{50}{25} \\
\mathrm{f}_{\mathrm{c}}=2 \mathrm{~cm}
\end{aligned}\)
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Ray Optics
282874
The magnifying power of a telescope is \(\mathrm{m}^{\prime}\). If the focal length of the eye piece is doubled, then its magnifying power will become
1 \(\sqrt{2} \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{\mathrm{m}}{2}\)
Explanation:
D: For a telescope, Magnifying power
\(m=\frac{\text { focal length of objective }\left(f_0\right)}{\text { focallength of eyepiece }\left(f_c\right)}\)
\(\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_e}\)
If, \(\quad \mathrm{f}_{\mathrm{s}}^{\prime}=2 \mathrm{f}_{\mathrm{c}}\)
New magnifying power,
\(\mathrm{m}^{\prime}=\frac{\mathrm{f}_o}{\mathrm{f}_e^{\prime}}=\frac{\mathrm{f}_{\mathrm{o}}}{2 \mathrm{f}_e}=\frac{\mathrm{m}}{2}\)
Manipal UGET -2020
Ray Optics
282875
Find the angular magnification of telescope if focal length of objective and eye lenses are 10 \(\mathrm{cm}\) and \(10 \mathrm{~mm}\) respectively and tube length is \(11 \mathrm{~cm}\) :
1 10
2 5
3 100
4 50
Explanation:
A: Given,
Focal length of object \(\left(f_{\mathrm{o}}\right)=10 \mathrm{~cm}\) and focal length of eye lens \(\left(\mathrm{f}_{\mathrm{c}}\right)=10 \mathrm{~mm}=1 \mathrm{~cm}\) As we know that,
Angular magnification of telescope \(=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_e}=\frac{10}{\mathrm{l}}=10\)
AIIMS-26.05.2019(E) Shift-2
Ray Optics
282876
The limit of resolution of a telescope is \(2.5 \times 10^{-}\) rad. If the telescope is used to detect light of wavelength \(500 \mathrm{~mm}\) coming from a star, the diameter of the objective lens used by telescope is
1 \(244 \mathrm{~cm}\)
2 \(258 \mathrm{~cm}\)
3 \(228 \mathrm{~cm}\)
4 \(264 \mathrm{~cm}\)
Explanation:
A: Given,
Limit of resolution \((\alpha)=2.5 \times 10^{-7}\) rad
Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\)
We know that,
Diameter of lens \((d)=\frac{1.22 \lambda}{\alpha}\)
\(\begin{aligned}
d=\frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-7}} \\
d=244 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-04.05.2019
Ray Optics
282877
The focal length of the objective lens of a telescope is \(50 \mathrm{~cm}\). If the magnification of the telescope is 25 , then the focal length of the eyepiece is
1 \(12.5 \mathrm{~cm}\)
2 \(5 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
C: Given,
Focal length of objective \(\left(f_0\right)=50 \mathrm{~cm}\)
Magnification \((\mathrm{m})=25\)
We know that.
Magnification of telescope,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}} \\
25=\frac{50}{\mathrm{f}_{\mathrm{e}}} \\
\mathrm{f}_{\mathrm{c}}=\frac{50}{25} \\
\mathrm{f}_{\mathrm{c}}=2 \mathrm{~cm}
\end{aligned}\)
282874
The magnifying power of a telescope is \(\mathrm{m}^{\prime}\). If the focal length of the eye piece is doubled, then its magnifying power will become
1 \(\sqrt{2} \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{\mathrm{m}}{2}\)
Explanation:
D: For a telescope, Magnifying power
\(m=\frac{\text { focal length of objective }\left(f_0\right)}{\text { focallength of eyepiece }\left(f_c\right)}\)
\(\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_e}\)
If, \(\quad \mathrm{f}_{\mathrm{s}}^{\prime}=2 \mathrm{f}_{\mathrm{c}}\)
New magnifying power,
\(\mathrm{m}^{\prime}=\frac{\mathrm{f}_o}{\mathrm{f}_e^{\prime}}=\frac{\mathrm{f}_{\mathrm{o}}}{2 \mathrm{f}_e}=\frac{\mathrm{m}}{2}\)
Manipal UGET -2020
Ray Optics
282875
Find the angular magnification of telescope if focal length of objective and eye lenses are 10 \(\mathrm{cm}\) and \(10 \mathrm{~mm}\) respectively and tube length is \(11 \mathrm{~cm}\) :
1 10
2 5
3 100
4 50
Explanation:
A: Given,
Focal length of object \(\left(f_{\mathrm{o}}\right)=10 \mathrm{~cm}\) and focal length of eye lens \(\left(\mathrm{f}_{\mathrm{c}}\right)=10 \mathrm{~mm}=1 \mathrm{~cm}\) As we know that,
Angular magnification of telescope \(=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_e}=\frac{10}{\mathrm{l}}=10\)
AIIMS-26.05.2019(E) Shift-2
Ray Optics
282876
The limit of resolution of a telescope is \(2.5 \times 10^{-}\) rad. If the telescope is used to detect light of wavelength \(500 \mathrm{~mm}\) coming from a star, the diameter of the objective lens used by telescope is
1 \(244 \mathrm{~cm}\)
2 \(258 \mathrm{~cm}\)
3 \(228 \mathrm{~cm}\)
4 \(264 \mathrm{~cm}\)
Explanation:
A: Given,
Limit of resolution \((\alpha)=2.5 \times 10^{-7}\) rad
Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\)
We know that,
Diameter of lens \((d)=\frac{1.22 \lambda}{\alpha}\)
\(\begin{aligned}
d=\frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-7}} \\
d=244 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-04.05.2019
Ray Optics
282877
The focal length of the objective lens of a telescope is \(50 \mathrm{~cm}\). If the magnification of the telescope is 25 , then the focal length of the eyepiece is
1 \(12.5 \mathrm{~cm}\)
2 \(5 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
C: Given,
Focal length of objective \(\left(f_0\right)=50 \mathrm{~cm}\)
Magnification \((\mathrm{m})=25\)
We know that.
Magnification of telescope,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}} \\
25=\frac{50}{\mathrm{f}_{\mathrm{e}}} \\
\mathrm{f}_{\mathrm{c}}=\frac{50}{25} \\
\mathrm{f}_{\mathrm{c}}=2 \mathrm{~cm}
\end{aligned}\)
282874
The magnifying power of a telescope is \(\mathrm{m}^{\prime}\). If the focal length of the eye piece is doubled, then its magnifying power will become
1 \(\sqrt{2} \mathrm{~m}\)
2 \(3 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{\mathrm{m}}{2}\)
Explanation:
D: For a telescope, Magnifying power
\(m=\frac{\text { focal length of objective }\left(f_0\right)}{\text { focallength of eyepiece }\left(f_c\right)}\)
\(\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_e}\)
If, \(\quad \mathrm{f}_{\mathrm{s}}^{\prime}=2 \mathrm{f}_{\mathrm{c}}\)
New magnifying power,
\(\mathrm{m}^{\prime}=\frac{\mathrm{f}_o}{\mathrm{f}_e^{\prime}}=\frac{\mathrm{f}_{\mathrm{o}}}{2 \mathrm{f}_e}=\frac{\mathrm{m}}{2}\)
Manipal UGET -2020
Ray Optics
282875
Find the angular magnification of telescope if focal length of objective and eye lenses are 10 \(\mathrm{cm}\) and \(10 \mathrm{~mm}\) respectively and tube length is \(11 \mathrm{~cm}\) :
1 10
2 5
3 100
4 50
Explanation:
A: Given,
Focal length of object \(\left(f_{\mathrm{o}}\right)=10 \mathrm{~cm}\) and focal length of eye lens \(\left(\mathrm{f}_{\mathrm{c}}\right)=10 \mathrm{~mm}=1 \mathrm{~cm}\) As we know that,
Angular magnification of telescope \(=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_e}=\frac{10}{\mathrm{l}}=10\)
AIIMS-26.05.2019(E) Shift-2
Ray Optics
282876
The limit of resolution of a telescope is \(2.5 \times 10^{-}\) rad. If the telescope is used to detect light of wavelength \(500 \mathrm{~mm}\) coming from a star, the diameter of the objective lens used by telescope is
1 \(244 \mathrm{~cm}\)
2 \(258 \mathrm{~cm}\)
3 \(228 \mathrm{~cm}\)
4 \(264 \mathrm{~cm}\)
Explanation:
A: Given,
Limit of resolution \((\alpha)=2.5 \times 10^{-7}\) rad
Wavelength \((\lambda)=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}\)
We know that,
Diameter of lens \((d)=\frac{1.22 \lambda}{\alpha}\)
\(\begin{aligned}
d=\frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-7}} \\
d=244 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-04.05.2019
Ray Optics
282877
The focal length of the objective lens of a telescope is \(50 \mathrm{~cm}\). If the magnification of the telescope is 25 , then the focal length of the eyepiece is
1 \(12.5 \mathrm{~cm}\)
2 \(5 \mathrm{~cm}\)
3 \(2 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
C: Given,
Focal length of objective \(\left(f_0\right)=50 \mathrm{~cm}\)
Magnification \((\mathrm{m})=25\)
We know that.
Magnification of telescope,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}} \\
25=\frac{50}{\mathrm{f}_{\mathrm{e}}} \\
\mathrm{f}_{\mathrm{c}}=\frac{50}{25} \\
\mathrm{f}_{\mathrm{c}}=2 \mathrm{~cm}
\end{aligned}\)