282884
For a telescope, focal length of objective lens is \(15 \mathrm{~cm}\) and focal length of eye piece is \(10 \mathrm{~mm}\). If tube length is \(16 \mathrm{~cm}\), then find the magnification in normal at adjustment.
1 150
2 15
3 1.5
4 10
Explanation:
B: Given that,
\(\mathrm{f}_{\mathrm{o}}=15 \mathrm{~cm}, \quad \mathrm{f}_{\mathrm{e}}=10 \mathrm{~mm}=1 \mathrm{~cm}\)
Magnification at normal adjustment -
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{1} \\
\mathrm{~m}=15
\end{aligned}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282885
The ratio of resolving power of telescope, when lights of wavelengths \(4400 \AA\) and \(5500 \AA\) are used, is
1 \(16: 25\)
2 \(4: 5\)
3 \(9: 1\)
4 \(5: 4\)
Explanation:
D: Given that,
\(\begin{aligned}
\lambda_1=4400 \AA \\
\lambda_2=5500 \AA
\end{aligned}\)
Ratio of resolving power,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\lambda_2}{\lambda_1} \quad\left(\because \mathrm{R} \propto \frac{1}{\lambda}\right)\)
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5500 \AA}{4400 \AA}\)
\(\mathrm{R}_1: \mathrm{R}_2=5: 4\)
GUJCET 2019
Ray Optics
282886
If the tube length of astronomical telescope is \(96 \mathrm{~cm}\) and magnifying power is 15 for normal setting, then the focal length of the objective is cm.
1 100
2 90
3 105
4 92
Explanation:
B: Given that,
Magnification, \(\mathrm{m}=15\)
\(\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=15\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{e}}=\frac{\mathrm{f}_{\mathrm{o}}}{15}\)
Also, given tube length is,
\(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=96 \mathrm{~cm}\)
\(\Rightarrow \mathrm{f}_{\mathrm{o}}+\frac{\mathrm{f}_{\mathrm{o}}}{15}=96 \mathrm{~cm} \Rightarrow \mathrm{f}_{\mathrm{o}}=\frac{96 \times 15}{16} \mathrm{~cm}\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{o}}=90 \mathrm{~cm}\)
GUJCET 2019
Ray Optics
282887
Magnifying power of an astronomical telescope for normal adjustment is \(\mathbf{1 0}\) and length of the telescope is \(110 \mathrm{~cm}\). Magnifying power of the same telescope, when the image is formed at the near point is
1 14
2 18
3 23
4 26
Explanation:
A: Given length of telescope, \(\mathrm{L}=110 \mathrm{~cm}\)
Magnification of normal adjustment \(\mathrm{m}_1=10\)
\(\begin{aligned}
\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=10 \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Also, \(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=110\)
\(\begin{aligned}
10 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=110 \\
\mathrm{f}_{\mathrm{e}}=\frac{110}{11}=10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}=10 \times 10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=100 \mathrm{~cm}
\end{aligned}\)
Magnifying power when image is at least distance of distinct vision-
\(\begin{aligned}
M_2=-f_o\left[\frac{1}{D}+\frac{1}{f_e}\right] \\
M_2=-100\left[\frac{1}{25}+\frac{1}{10}\right] \\
M_2=-14 \quad \text { \{only magnitude\} }
\end{aligned}\)
282884
For a telescope, focal length of objective lens is \(15 \mathrm{~cm}\) and focal length of eye piece is \(10 \mathrm{~mm}\). If tube length is \(16 \mathrm{~cm}\), then find the magnification in normal at adjustment.
1 150
2 15
3 1.5
4 10
Explanation:
B: Given that,
\(\mathrm{f}_{\mathrm{o}}=15 \mathrm{~cm}, \quad \mathrm{f}_{\mathrm{e}}=10 \mathrm{~mm}=1 \mathrm{~cm}\)
Magnification at normal adjustment -
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{1} \\
\mathrm{~m}=15
\end{aligned}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282885
The ratio of resolving power of telescope, when lights of wavelengths \(4400 \AA\) and \(5500 \AA\) are used, is
1 \(16: 25\)
2 \(4: 5\)
3 \(9: 1\)
4 \(5: 4\)
Explanation:
D: Given that,
\(\begin{aligned}
\lambda_1=4400 \AA \\
\lambda_2=5500 \AA
\end{aligned}\)
Ratio of resolving power,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\lambda_2}{\lambda_1} \quad\left(\because \mathrm{R} \propto \frac{1}{\lambda}\right)\)
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5500 \AA}{4400 \AA}\)
\(\mathrm{R}_1: \mathrm{R}_2=5: 4\)
GUJCET 2019
Ray Optics
282886
If the tube length of astronomical telescope is \(96 \mathrm{~cm}\) and magnifying power is 15 for normal setting, then the focal length of the objective is cm.
1 100
2 90
3 105
4 92
Explanation:
B: Given that,
Magnification, \(\mathrm{m}=15\)
\(\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=15\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{e}}=\frac{\mathrm{f}_{\mathrm{o}}}{15}\)
Also, given tube length is,
\(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=96 \mathrm{~cm}\)
\(\Rightarrow \mathrm{f}_{\mathrm{o}}+\frac{\mathrm{f}_{\mathrm{o}}}{15}=96 \mathrm{~cm} \Rightarrow \mathrm{f}_{\mathrm{o}}=\frac{96 \times 15}{16} \mathrm{~cm}\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{o}}=90 \mathrm{~cm}\)
GUJCET 2019
Ray Optics
282887
Magnifying power of an astronomical telescope for normal adjustment is \(\mathbf{1 0}\) and length of the telescope is \(110 \mathrm{~cm}\). Magnifying power of the same telescope, when the image is formed at the near point is
1 14
2 18
3 23
4 26
Explanation:
A: Given length of telescope, \(\mathrm{L}=110 \mathrm{~cm}\)
Magnification of normal adjustment \(\mathrm{m}_1=10\)
\(\begin{aligned}
\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=10 \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Also, \(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=110\)
\(\begin{aligned}
10 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=110 \\
\mathrm{f}_{\mathrm{e}}=\frac{110}{11}=10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}=10 \times 10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=100 \mathrm{~cm}
\end{aligned}\)
Magnifying power when image is at least distance of distinct vision-
\(\begin{aligned}
M_2=-f_o\left[\frac{1}{D}+\frac{1}{f_e}\right] \\
M_2=-100\left[\frac{1}{25}+\frac{1}{10}\right] \\
M_2=-14 \quad \text { \{only magnitude\} }
\end{aligned}\)
282884
For a telescope, focal length of objective lens is \(15 \mathrm{~cm}\) and focal length of eye piece is \(10 \mathrm{~mm}\). If tube length is \(16 \mathrm{~cm}\), then find the magnification in normal at adjustment.
1 150
2 15
3 1.5
4 10
Explanation:
B: Given that,
\(\mathrm{f}_{\mathrm{o}}=15 \mathrm{~cm}, \quad \mathrm{f}_{\mathrm{e}}=10 \mathrm{~mm}=1 \mathrm{~cm}\)
Magnification at normal adjustment -
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{1} \\
\mathrm{~m}=15
\end{aligned}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282885
The ratio of resolving power of telescope, when lights of wavelengths \(4400 \AA\) and \(5500 \AA\) are used, is
1 \(16: 25\)
2 \(4: 5\)
3 \(9: 1\)
4 \(5: 4\)
Explanation:
D: Given that,
\(\begin{aligned}
\lambda_1=4400 \AA \\
\lambda_2=5500 \AA
\end{aligned}\)
Ratio of resolving power,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\lambda_2}{\lambda_1} \quad\left(\because \mathrm{R} \propto \frac{1}{\lambda}\right)\)
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5500 \AA}{4400 \AA}\)
\(\mathrm{R}_1: \mathrm{R}_2=5: 4\)
GUJCET 2019
Ray Optics
282886
If the tube length of astronomical telescope is \(96 \mathrm{~cm}\) and magnifying power is 15 for normal setting, then the focal length of the objective is cm.
1 100
2 90
3 105
4 92
Explanation:
B: Given that,
Magnification, \(\mathrm{m}=15\)
\(\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=15\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{e}}=\frac{\mathrm{f}_{\mathrm{o}}}{15}\)
Also, given tube length is,
\(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=96 \mathrm{~cm}\)
\(\Rightarrow \mathrm{f}_{\mathrm{o}}+\frac{\mathrm{f}_{\mathrm{o}}}{15}=96 \mathrm{~cm} \Rightarrow \mathrm{f}_{\mathrm{o}}=\frac{96 \times 15}{16} \mathrm{~cm}\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{o}}=90 \mathrm{~cm}\)
GUJCET 2019
Ray Optics
282887
Magnifying power of an astronomical telescope for normal adjustment is \(\mathbf{1 0}\) and length of the telescope is \(110 \mathrm{~cm}\). Magnifying power of the same telescope, when the image is formed at the near point is
1 14
2 18
3 23
4 26
Explanation:
A: Given length of telescope, \(\mathrm{L}=110 \mathrm{~cm}\)
Magnification of normal adjustment \(\mathrm{m}_1=10\)
\(\begin{aligned}
\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=10 \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Also, \(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=110\)
\(\begin{aligned}
10 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=110 \\
\mathrm{f}_{\mathrm{e}}=\frac{110}{11}=10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}=10 \times 10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=100 \mathrm{~cm}
\end{aligned}\)
Magnifying power when image is at least distance of distinct vision-
\(\begin{aligned}
M_2=-f_o\left[\frac{1}{D}+\frac{1}{f_e}\right] \\
M_2=-100\left[\frac{1}{25}+\frac{1}{10}\right] \\
M_2=-14 \quad \text { \{only magnitude\} }
\end{aligned}\)
282884
For a telescope, focal length of objective lens is \(15 \mathrm{~cm}\) and focal length of eye piece is \(10 \mathrm{~mm}\). If tube length is \(16 \mathrm{~cm}\), then find the magnification in normal at adjustment.
1 150
2 15
3 1.5
4 10
Explanation:
B: Given that,
\(\mathrm{f}_{\mathrm{o}}=15 \mathrm{~cm}, \quad \mathrm{f}_{\mathrm{e}}=10 \mathrm{~mm}=1 \mathrm{~cm}\)
Magnification at normal adjustment -
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{1} \\
\mathrm{~m}=15
\end{aligned}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282885
The ratio of resolving power of telescope, when lights of wavelengths \(4400 \AA\) and \(5500 \AA\) are used, is
1 \(16: 25\)
2 \(4: 5\)
3 \(9: 1\)
4 \(5: 4\)
Explanation:
D: Given that,
\(\begin{aligned}
\lambda_1=4400 \AA \\
\lambda_2=5500 \AA
\end{aligned}\)
Ratio of resolving power,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\lambda_2}{\lambda_1} \quad\left(\because \mathrm{R} \propto \frac{1}{\lambda}\right)\)
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5500 \AA}{4400 \AA}\)
\(\mathrm{R}_1: \mathrm{R}_2=5: 4\)
GUJCET 2019
Ray Optics
282886
If the tube length of astronomical telescope is \(96 \mathrm{~cm}\) and magnifying power is 15 for normal setting, then the focal length of the objective is cm.
1 100
2 90
3 105
4 92
Explanation:
B: Given that,
Magnification, \(\mathrm{m}=15\)
\(\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=15\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{e}}=\frac{\mathrm{f}_{\mathrm{o}}}{15}\)
Also, given tube length is,
\(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=96 \mathrm{~cm}\)
\(\Rightarrow \mathrm{f}_{\mathrm{o}}+\frac{\mathrm{f}_{\mathrm{o}}}{15}=96 \mathrm{~cm} \Rightarrow \mathrm{f}_{\mathrm{o}}=\frac{96 \times 15}{16} \mathrm{~cm}\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{o}}=90 \mathrm{~cm}\)
GUJCET 2019
Ray Optics
282887
Magnifying power of an astronomical telescope for normal adjustment is \(\mathbf{1 0}\) and length of the telescope is \(110 \mathrm{~cm}\). Magnifying power of the same telescope, when the image is formed at the near point is
1 14
2 18
3 23
4 26
Explanation:
A: Given length of telescope, \(\mathrm{L}=110 \mathrm{~cm}\)
Magnification of normal adjustment \(\mathrm{m}_1=10\)
\(\begin{aligned}
\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=10 \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Also, \(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=110\)
\(\begin{aligned}
10 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=110 \\
\mathrm{f}_{\mathrm{e}}=\frac{110}{11}=10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}=10 \times 10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=100 \mathrm{~cm}
\end{aligned}\)
Magnifying power when image is at least distance of distinct vision-
\(\begin{aligned}
M_2=-f_o\left[\frac{1}{D}+\frac{1}{f_e}\right] \\
M_2=-100\left[\frac{1}{25}+\frac{1}{10}\right] \\
M_2=-14 \quad \text { \{only magnitude\} }
\end{aligned}\)
282884
For a telescope, focal length of objective lens is \(15 \mathrm{~cm}\) and focal length of eye piece is \(10 \mathrm{~mm}\). If tube length is \(16 \mathrm{~cm}\), then find the magnification in normal at adjustment.
1 150
2 15
3 1.5
4 10
Explanation:
B: Given that,
\(\mathrm{f}_{\mathrm{o}}=15 \mathrm{~cm}, \quad \mathrm{f}_{\mathrm{e}}=10 \mathrm{~mm}=1 \mathrm{~cm}\)
Magnification at normal adjustment -
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{1} \\
\mathrm{~m}=15
\end{aligned}\)
AIIMS-25.05.2019(M) Shift-1
Ray Optics
282885
The ratio of resolving power of telescope, when lights of wavelengths \(4400 \AA\) and \(5500 \AA\) are used, is
1 \(16: 25\)
2 \(4: 5\)
3 \(9: 1\)
4 \(5: 4\)
Explanation:
D: Given that,
\(\begin{aligned}
\lambda_1=4400 \AA \\
\lambda_2=5500 \AA
\end{aligned}\)
Ratio of resolving power,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\lambda_2}{\lambda_1} \quad\left(\because \mathrm{R} \propto \frac{1}{\lambda}\right)\)
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5500 \AA}{4400 \AA}\)
\(\mathrm{R}_1: \mathrm{R}_2=5: 4\)
GUJCET 2019
Ray Optics
282886
If the tube length of astronomical telescope is \(96 \mathrm{~cm}\) and magnifying power is 15 for normal setting, then the focal length of the objective is cm.
1 100
2 90
3 105
4 92
Explanation:
B: Given that,
Magnification, \(\mathrm{m}=15\)
\(\mathrm{m}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=15\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{e}}=\frac{\mathrm{f}_{\mathrm{o}}}{15}\)
Also, given tube length is,
\(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=96 \mathrm{~cm}\)
\(\Rightarrow \mathrm{f}_{\mathrm{o}}+\frac{\mathrm{f}_{\mathrm{o}}}{15}=96 \mathrm{~cm} \Rightarrow \mathrm{f}_{\mathrm{o}}=\frac{96 \times 15}{16} \mathrm{~cm}\)
\(\Rightarrow \quad \mathrm{f}_{\mathrm{o}}=90 \mathrm{~cm}\)
GUJCET 2019
Ray Optics
282887
Magnifying power of an astronomical telescope for normal adjustment is \(\mathbf{1 0}\) and length of the telescope is \(110 \mathrm{~cm}\). Magnifying power of the same telescope, when the image is formed at the near point is
1 14
2 18
3 23
4 26
Explanation:
A: Given length of telescope, \(\mathrm{L}=110 \mathrm{~cm}\)
Magnification of normal adjustment \(\mathrm{m}_1=10\)
\(\begin{aligned}
\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=10 \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Also, \(\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=110\)
\(\begin{aligned}
10 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=110 \\
\mathrm{f}_{\mathrm{e}}=\frac{110}{11}=10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=10 \mathrm{f}_{\mathrm{e}}=10 \times 10 \mathrm{~cm} \\
\mathrm{f}_{\mathrm{o}}=100 \mathrm{~cm}
\end{aligned}\)
Magnifying power when image is at least distance of distinct vision-
\(\begin{aligned}
M_2=-f_o\left[\frac{1}{D}+\frac{1}{f_e}\right] \\
M_2=-100\left[\frac{1}{25}+\frac{1}{10}\right] \\
M_2=-14 \quad \text { \{only magnitude\} }
\end{aligned}\)