282878
If \(\mathrm{f}_0=5 \mathrm{~cm}, \lambda=6000 \stackrel{\circ}{\mathrm{A}}, \mathrm{a}=1 \mathrm{~cm}\), for \(\mathrm{a}\) microscope, then what will be its resolving power?
1 \(11.9 \times 10^5 / \mathrm{m}\)
2 \(10.9 \times 10^5 / \mathrm{m}\)
3 \(10.9 \times 10^4 / \mathrm{m}\)
4 \(10.9 \times 10^3 / \mathrm{m}\)
Explanation:
B: Given that,
Wavelength of light \((\lambda)=6000 \AA\)
\(\mathrm{f}_0=5 \mathrm{~cm}, \mathrm{a}=1 \mathrm{~cm}\)
\(\therefore \quad\) Resolving power \(=\frac{2 \mu \sin \theta}{1.22 \lambda}\)
From figure -
If \(\theta\) is small,
\(\tan \theta \square \sin \theta=\frac{\mathrm{a}}{\mathrm{f}_0}=\frac{1}{5}=0.2\)
\(\because \quad\) R.P \(=\frac{2 \times 1 \times 0.2}{1.22 \times 6 \times 10^{-7}}\)
\([\because \mu=1]\)
R.P. \(=\frac{4 \times 10^6}{3.66}\)
R. \(\mathrm{P}=10.9 \times 10^5 / \mathrm{m}\)
AIIMS-26.05.2019(M) Shift-1
Ray Optics
282879
'Lumen' is the unit of
1 Luminous flux
2 Luminous intensity
3 Illuminance
4 Light frequency
Explanation:
A: Lumen is the unit of luminous flux.
Luminous flux or amount of light is defined as the amount of streaming outward through on steradian (a unit of solid angle), from a uniform point source having an intensity of one candela.
UPSEE 2019
Ray Optics
282880
The resolving power of telescope can be increased by:
1 increasing the diameter of object
2 increasing the wavelength of light used
3 decreasing the diameter of objective
4 decreasing the frequency of light used
Explanation:
A: Resolving power of telescope \((\mathrm{R})=\frac{\mathrm{a}}{1.22 \lambda}\)
Where, \(\mathrm{a}=\) Diameter of objective lens of telescope \(\mathrm{R} \propto \mathrm{a}\)
So, the resolving power of telescope can be increased by increasing the diameter of object.
UPSEE 2019
Ray Optics
282882
The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is \(20 \mathrm{~cm}\). The focal length of objective and eyepiece are respectively
1 \(10 \mathrm{~cm}, 10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
4 \(11 \mathrm{~cm}, 9 \mathrm{~cm}\)
Explanation:
C: Given,
Magnification \((\mathrm{m})=9\)
Length \((\mathrm{L})=20 \mathrm{~cm}\)
\(\begin{aligned}
m=\frac{f_o}{f_e}=9 \\
f_o=9 f_e
\end{aligned}\)
Also,
\(\begin{aligned}
\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=20 \\
9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20 \\
\mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}
\end{aligned}\)
And,
\(\begin{aligned}
\mathrm{f}_{\mathrm{o}}=9 \mathrm{f}_{\mathrm{e}}=9 \times 2 \\
\mathrm{f}_{\mathrm{o}}=18 \mathrm{~cm}
\end{aligned}\)
So, the focal length of objective and eyepiece are respectively \(18 \mathrm{~cm}\) and \(2 \mathrm{~cm}\).
282878
If \(\mathrm{f}_0=5 \mathrm{~cm}, \lambda=6000 \stackrel{\circ}{\mathrm{A}}, \mathrm{a}=1 \mathrm{~cm}\), for \(\mathrm{a}\) microscope, then what will be its resolving power?
1 \(11.9 \times 10^5 / \mathrm{m}\)
2 \(10.9 \times 10^5 / \mathrm{m}\)
3 \(10.9 \times 10^4 / \mathrm{m}\)
4 \(10.9 \times 10^3 / \mathrm{m}\)
Explanation:
B: Given that,
Wavelength of light \((\lambda)=6000 \AA\)
\(\mathrm{f}_0=5 \mathrm{~cm}, \mathrm{a}=1 \mathrm{~cm}\)
\(\therefore \quad\) Resolving power \(=\frac{2 \mu \sin \theta}{1.22 \lambda}\)
From figure -
If \(\theta\) is small,
\(\tan \theta \square \sin \theta=\frac{\mathrm{a}}{\mathrm{f}_0}=\frac{1}{5}=0.2\)
\(\because \quad\) R.P \(=\frac{2 \times 1 \times 0.2}{1.22 \times 6 \times 10^{-7}}\)
\([\because \mu=1]\)
R.P. \(=\frac{4 \times 10^6}{3.66}\)
R. \(\mathrm{P}=10.9 \times 10^5 / \mathrm{m}\)
AIIMS-26.05.2019(M) Shift-1
Ray Optics
282879
'Lumen' is the unit of
1 Luminous flux
2 Luminous intensity
3 Illuminance
4 Light frequency
Explanation:
A: Lumen is the unit of luminous flux.
Luminous flux or amount of light is defined as the amount of streaming outward through on steradian (a unit of solid angle), from a uniform point source having an intensity of one candela.
UPSEE 2019
Ray Optics
282880
The resolving power of telescope can be increased by:
1 increasing the diameter of object
2 increasing the wavelength of light used
3 decreasing the diameter of objective
4 decreasing the frequency of light used
Explanation:
A: Resolving power of telescope \((\mathrm{R})=\frac{\mathrm{a}}{1.22 \lambda}\)
Where, \(\mathrm{a}=\) Diameter of objective lens of telescope \(\mathrm{R} \propto \mathrm{a}\)
So, the resolving power of telescope can be increased by increasing the diameter of object.
UPSEE 2019
Ray Optics
282882
The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is \(20 \mathrm{~cm}\). The focal length of objective and eyepiece are respectively
1 \(10 \mathrm{~cm}, 10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
4 \(11 \mathrm{~cm}, 9 \mathrm{~cm}\)
Explanation:
C: Given,
Magnification \((\mathrm{m})=9\)
Length \((\mathrm{L})=20 \mathrm{~cm}\)
\(\begin{aligned}
m=\frac{f_o}{f_e}=9 \\
f_o=9 f_e
\end{aligned}\)
Also,
\(\begin{aligned}
\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=20 \\
9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20 \\
\mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}
\end{aligned}\)
And,
\(\begin{aligned}
\mathrm{f}_{\mathrm{o}}=9 \mathrm{f}_{\mathrm{e}}=9 \times 2 \\
\mathrm{f}_{\mathrm{o}}=18 \mathrm{~cm}
\end{aligned}\)
So, the focal length of objective and eyepiece are respectively \(18 \mathrm{~cm}\) and \(2 \mathrm{~cm}\).
282878
If \(\mathrm{f}_0=5 \mathrm{~cm}, \lambda=6000 \stackrel{\circ}{\mathrm{A}}, \mathrm{a}=1 \mathrm{~cm}\), for \(\mathrm{a}\) microscope, then what will be its resolving power?
1 \(11.9 \times 10^5 / \mathrm{m}\)
2 \(10.9 \times 10^5 / \mathrm{m}\)
3 \(10.9 \times 10^4 / \mathrm{m}\)
4 \(10.9 \times 10^3 / \mathrm{m}\)
Explanation:
B: Given that,
Wavelength of light \((\lambda)=6000 \AA\)
\(\mathrm{f}_0=5 \mathrm{~cm}, \mathrm{a}=1 \mathrm{~cm}\)
\(\therefore \quad\) Resolving power \(=\frac{2 \mu \sin \theta}{1.22 \lambda}\)
From figure -
If \(\theta\) is small,
\(\tan \theta \square \sin \theta=\frac{\mathrm{a}}{\mathrm{f}_0}=\frac{1}{5}=0.2\)
\(\because \quad\) R.P \(=\frac{2 \times 1 \times 0.2}{1.22 \times 6 \times 10^{-7}}\)
\([\because \mu=1]\)
R.P. \(=\frac{4 \times 10^6}{3.66}\)
R. \(\mathrm{P}=10.9 \times 10^5 / \mathrm{m}\)
AIIMS-26.05.2019(M) Shift-1
Ray Optics
282879
'Lumen' is the unit of
1 Luminous flux
2 Luminous intensity
3 Illuminance
4 Light frequency
Explanation:
A: Lumen is the unit of luminous flux.
Luminous flux or amount of light is defined as the amount of streaming outward through on steradian (a unit of solid angle), from a uniform point source having an intensity of one candela.
UPSEE 2019
Ray Optics
282880
The resolving power of telescope can be increased by:
1 increasing the diameter of object
2 increasing the wavelength of light used
3 decreasing the diameter of objective
4 decreasing the frequency of light used
Explanation:
A: Resolving power of telescope \((\mathrm{R})=\frac{\mathrm{a}}{1.22 \lambda}\)
Where, \(\mathrm{a}=\) Diameter of objective lens of telescope \(\mathrm{R} \propto \mathrm{a}\)
So, the resolving power of telescope can be increased by increasing the diameter of object.
UPSEE 2019
Ray Optics
282882
The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is \(20 \mathrm{~cm}\). The focal length of objective and eyepiece are respectively
1 \(10 \mathrm{~cm}, 10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
4 \(11 \mathrm{~cm}, 9 \mathrm{~cm}\)
Explanation:
C: Given,
Magnification \((\mathrm{m})=9\)
Length \((\mathrm{L})=20 \mathrm{~cm}\)
\(\begin{aligned}
m=\frac{f_o}{f_e}=9 \\
f_o=9 f_e
\end{aligned}\)
Also,
\(\begin{aligned}
\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=20 \\
9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20 \\
\mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}
\end{aligned}\)
And,
\(\begin{aligned}
\mathrm{f}_{\mathrm{o}}=9 \mathrm{f}_{\mathrm{e}}=9 \times 2 \\
\mathrm{f}_{\mathrm{o}}=18 \mathrm{~cm}
\end{aligned}\)
So, the focal length of objective and eyepiece are respectively \(18 \mathrm{~cm}\) and \(2 \mathrm{~cm}\).
282878
If \(\mathrm{f}_0=5 \mathrm{~cm}, \lambda=6000 \stackrel{\circ}{\mathrm{A}}, \mathrm{a}=1 \mathrm{~cm}\), for \(\mathrm{a}\) microscope, then what will be its resolving power?
1 \(11.9 \times 10^5 / \mathrm{m}\)
2 \(10.9 \times 10^5 / \mathrm{m}\)
3 \(10.9 \times 10^4 / \mathrm{m}\)
4 \(10.9 \times 10^3 / \mathrm{m}\)
Explanation:
B: Given that,
Wavelength of light \((\lambda)=6000 \AA\)
\(\mathrm{f}_0=5 \mathrm{~cm}, \mathrm{a}=1 \mathrm{~cm}\)
\(\therefore \quad\) Resolving power \(=\frac{2 \mu \sin \theta}{1.22 \lambda}\)
From figure -
If \(\theta\) is small,
\(\tan \theta \square \sin \theta=\frac{\mathrm{a}}{\mathrm{f}_0}=\frac{1}{5}=0.2\)
\(\because \quad\) R.P \(=\frac{2 \times 1 \times 0.2}{1.22 \times 6 \times 10^{-7}}\)
\([\because \mu=1]\)
R.P. \(=\frac{4 \times 10^6}{3.66}\)
R. \(\mathrm{P}=10.9 \times 10^5 / \mathrm{m}\)
AIIMS-26.05.2019(M) Shift-1
Ray Optics
282879
'Lumen' is the unit of
1 Luminous flux
2 Luminous intensity
3 Illuminance
4 Light frequency
Explanation:
A: Lumen is the unit of luminous flux.
Luminous flux or amount of light is defined as the amount of streaming outward through on steradian (a unit of solid angle), from a uniform point source having an intensity of one candela.
UPSEE 2019
Ray Optics
282880
The resolving power of telescope can be increased by:
1 increasing the diameter of object
2 increasing the wavelength of light used
3 decreasing the diameter of objective
4 decreasing the frequency of light used
Explanation:
A: Resolving power of telescope \((\mathrm{R})=\frac{\mathrm{a}}{1.22 \lambda}\)
Where, \(\mathrm{a}=\) Diameter of objective lens of telescope \(\mathrm{R} \propto \mathrm{a}\)
So, the resolving power of telescope can be increased by increasing the diameter of object.
UPSEE 2019
Ray Optics
282882
The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is \(20 \mathrm{~cm}\). The focal length of objective and eyepiece are respectively
1 \(10 \mathrm{~cm}, 10 \mathrm{~cm}\)
2 \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)
3 \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
4 \(11 \mathrm{~cm}, 9 \mathrm{~cm}\)
Explanation:
C: Given,
Magnification \((\mathrm{m})=9\)
Length \((\mathrm{L})=20 \mathrm{~cm}\)
\(\begin{aligned}
m=\frac{f_o}{f_e}=9 \\
f_o=9 f_e
\end{aligned}\)
Also,
\(\begin{aligned}
\mathrm{L}=\mathrm{f}_{\mathrm{o}}+\mathrm{f}_{\mathrm{e}}=20 \\
9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20 \\
\mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}
\end{aligned}\)
And,
\(\begin{aligned}
\mathrm{f}_{\mathrm{o}}=9 \mathrm{f}_{\mathrm{e}}=9 \times 2 \\
\mathrm{f}_{\mathrm{o}}=18 \mathrm{~cm}
\end{aligned}\)
So, the focal length of objective and eyepiece are respectively \(18 \mathrm{~cm}\) and \(2 \mathrm{~cm}\).
