NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282351
When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:
1 dispersion of light
2 reflection of light
3 polarization of light
4 interference of light
Explanation:
D Interference of light -: When two light waves from different coherent sources meet together, then the distribution of energy due to one wave is distributed by other.
This modification of the distribution of light energy due to super position of two light waves is called "interference of light".
Hence, the colour in day light on a drop of oil is due to inter ference of light.
Assam CEE-2017
Ray Optics
282352
The plane face of a Plano convex lens is silvered. If \(\mu\) be the refractive index and \(R\) is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature
1 \(\mu \mathrm{R}\)
2 \(\mathrm{R}^2 / \mu\)
3 \(\mathrm{R} /(\mu-1)\)
4 \([(\mu+1) /(\mu-1)] \mathrm{R}\)
Explanation:
C: Consider a parallel beam of light incident on convex surface,
\(\begin{aligned}
\frac{\mu}{\mathrm{v}_1}-\frac{1}{-\infty}=\frac{\mu-1}{\mathrm{R}} \\
\mathrm{v}_1=\frac{\mu \mathrm{R}}{\mu-1}
\end{aligned}\)
This image gets reflected from the mirror to form an image at distance \(\left\{\frac{\mu \mathrm{R}}{\mu-1}\right\}\) to the left of lens.
Refraction again from convex surface-
\(\begin{aligned}
\frac{1}{V_2}-\frac{\mu}{\frac{\mu R}{\mu-1}}=\frac{1-\mu}{-R} \\
\frac{1}{V_2}=\frac{2(\mu-1)}{R} \\
\frac{1}{f}=\frac{2(\mu-1)}{R} \\
f=\frac{R}{2(\mu-1)} \\
R^{\prime}=2 f=\frac{R}{\mu-1}
\end{aligned}\)
(left of side)
Hence, \(R^{\prime}=\frac{R}{\mu-1}\)
JIPMER-2017
Ray Optics
282353
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is \(30 \mathrm{~cm}\). The focal length of the objective, when the angular magnification of the telescope is 2 , will be:
1 \(20 \mathrm{~cm}\)
2 \(30 \mathrm{~cm}\)
3 \(10 \mathrm{~cm}\)
4 \(15 \mathrm{~cm}\)
Explanation:
AGiven, \(\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=30 \mathrm{~cm}\)
And,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=2 \\
\mathrm{f}_0=2 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Putting the value \(\mathrm{f}_0\) in equation (i), we get-
\(\begin{aligned}
2 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=30 \\
\mathrm{f}_{\mathrm{e}}=\frac{30}{3}=10 \mathrm{~cm}
\end{aligned}\)
Then,
\(\mathrm{f}_0=2 \mathrm{f}_{\mathrm{c}}=2 \times 10\)
So, \(\quad \mathrm{f}_0=20 \mathrm{~cm}\)
JEE Main-28.07.2022
Ray Optics
282354
The power of a lens (biconvex) is \(1.25 \mathrm{~m}^{-1}\) in particular medium. Refractive index of the lens is \(\mathbf{1 . 5}\) and radii of curvature are \(20 \mathrm{~cm}\) and 40 cm respectively. The refractive index of surrounding medium:
282351
When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:
1 dispersion of light
2 reflection of light
3 polarization of light
4 interference of light
Explanation:
D Interference of light -: When two light waves from different coherent sources meet together, then the distribution of energy due to one wave is distributed by other.
This modification of the distribution of light energy due to super position of two light waves is called "interference of light".
Hence, the colour in day light on a drop of oil is due to inter ference of light.
Assam CEE-2017
Ray Optics
282352
The plane face of a Plano convex lens is silvered. If \(\mu\) be the refractive index and \(R\) is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature
1 \(\mu \mathrm{R}\)
2 \(\mathrm{R}^2 / \mu\)
3 \(\mathrm{R} /(\mu-1)\)
4 \([(\mu+1) /(\mu-1)] \mathrm{R}\)
Explanation:
C: Consider a parallel beam of light incident on convex surface,
\(\begin{aligned}
\frac{\mu}{\mathrm{v}_1}-\frac{1}{-\infty}=\frac{\mu-1}{\mathrm{R}} \\
\mathrm{v}_1=\frac{\mu \mathrm{R}}{\mu-1}
\end{aligned}\)
This image gets reflected from the mirror to form an image at distance \(\left\{\frac{\mu \mathrm{R}}{\mu-1}\right\}\) to the left of lens.
Refraction again from convex surface-
\(\begin{aligned}
\frac{1}{V_2}-\frac{\mu}{\frac{\mu R}{\mu-1}}=\frac{1-\mu}{-R} \\
\frac{1}{V_2}=\frac{2(\mu-1)}{R} \\
\frac{1}{f}=\frac{2(\mu-1)}{R} \\
f=\frac{R}{2(\mu-1)} \\
R^{\prime}=2 f=\frac{R}{\mu-1}
\end{aligned}\)
(left of side)
Hence, \(R^{\prime}=\frac{R}{\mu-1}\)
JIPMER-2017
Ray Optics
282353
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is \(30 \mathrm{~cm}\). The focal length of the objective, when the angular magnification of the telescope is 2 , will be:
1 \(20 \mathrm{~cm}\)
2 \(30 \mathrm{~cm}\)
3 \(10 \mathrm{~cm}\)
4 \(15 \mathrm{~cm}\)
Explanation:
AGiven, \(\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=30 \mathrm{~cm}\)
And,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=2 \\
\mathrm{f}_0=2 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Putting the value \(\mathrm{f}_0\) in equation (i), we get-
\(\begin{aligned}
2 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=30 \\
\mathrm{f}_{\mathrm{e}}=\frac{30}{3}=10 \mathrm{~cm}
\end{aligned}\)
Then,
\(\mathrm{f}_0=2 \mathrm{f}_{\mathrm{c}}=2 \times 10\)
So, \(\quad \mathrm{f}_0=20 \mathrm{~cm}\)
JEE Main-28.07.2022
Ray Optics
282354
The power of a lens (biconvex) is \(1.25 \mathrm{~m}^{-1}\) in particular medium. Refractive index of the lens is \(\mathbf{1 . 5}\) and radii of curvature are \(20 \mathrm{~cm}\) and 40 cm respectively. The refractive index of surrounding medium:
282351
When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:
1 dispersion of light
2 reflection of light
3 polarization of light
4 interference of light
Explanation:
D Interference of light -: When two light waves from different coherent sources meet together, then the distribution of energy due to one wave is distributed by other.
This modification of the distribution of light energy due to super position of two light waves is called "interference of light".
Hence, the colour in day light on a drop of oil is due to inter ference of light.
Assam CEE-2017
Ray Optics
282352
The plane face of a Plano convex lens is silvered. If \(\mu\) be the refractive index and \(R\) is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature
1 \(\mu \mathrm{R}\)
2 \(\mathrm{R}^2 / \mu\)
3 \(\mathrm{R} /(\mu-1)\)
4 \([(\mu+1) /(\mu-1)] \mathrm{R}\)
Explanation:
C: Consider a parallel beam of light incident on convex surface,
\(\begin{aligned}
\frac{\mu}{\mathrm{v}_1}-\frac{1}{-\infty}=\frac{\mu-1}{\mathrm{R}} \\
\mathrm{v}_1=\frac{\mu \mathrm{R}}{\mu-1}
\end{aligned}\)
This image gets reflected from the mirror to form an image at distance \(\left\{\frac{\mu \mathrm{R}}{\mu-1}\right\}\) to the left of lens.
Refraction again from convex surface-
\(\begin{aligned}
\frac{1}{V_2}-\frac{\mu}{\frac{\mu R}{\mu-1}}=\frac{1-\mu}{-R} \\
\frac{1}{V_2}=\frac{2(\mu-1)}{R} \\
\frac{1}{f}=\frac{2(\mu-1)}{R} \\
f=\frac{R}{2(\mu-1)} \\
R^{\prime}=2 f=\frac{R}{\mu-1}
\end{aligned}\)
(left of side)
Hence, \(R^{\prime}=\frac{R}{\mu-1}\)
JIPMER-2017
Ray Optics
282353
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is \(30 \mathrm{~cm}\). The focal length of the objective, when the angular magnification of the telescope is 2 , will be:
1 \(20 \mathrm{~cm}\)
2 \(30 \mathrm{~cm}\)
3 \(10 \mathrm{~cm}\)
4 \(15 \mathrm{~cm}\)
Explanation:
AGiven, \(\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=30 \mathrm{~cm}\)
And,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=2 \\
\mathrm{f}_0=2 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Putting the value \(\mathrm{f}_0\) in equation (i), we get-
\(\begin{aligned}
2 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=30 \\
\mathrm{f}_{\mathrm{e}}=\frac{30}{3}=10 \mathrm{~cm}
\end{aligned}\)
Then,
\(\mathrm{f}_0=2 \mathrm{f}_{\mathrm{c}}=2 \times 10\)
So, \(\quad \mathrm{f}_0=20 \mathrm{~cm}\)
JEE Main-28.07.2022
Ray Optics
282354
The power of a lens (biconvex) is \(1.25 \mathrm{~m}^{-1}\) in particular medium. Refractive index of the lens is \(\mathbf{1 . 5}\) and radii of curvature are \(20 \mathrm{~cm}\) and 40 cm respectively. The refractive index of surrounding medium:
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ray Optics
282351
When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:
1 dispersion of light
2 reflection of light
3 polarization of light
4 interference of light
Explanation:
D Interference of light -: When two light waves from different coherent sources meet together, then the distribution of energy due to one wave is distributed by other.
This modification of the distribution of light energy due to super position of two light waves is called "interference of light".
Hence, the colour in day light on a drop of oil is due to inter ference of light.
Assam CEE-2017
Ray Optics
282352
The plane face of a Plano convex lens is silvered. If \(\mu\) be the refractive index and \(R\) is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature
1 \(\mu \mathrm{R}\)
2 \(\mathrm{R}^2 / \mu\)
3 \(\mathrm{R} /(\mu-1)\)
4 \([(\mu+1) /(\mu-1)] \mathrm{R}\)
Explanation:
C: Consider a parallel beam of light incident on convex surface,
\(\begin{aligned}
\frac{\mu}{\mathrm{v}_1}-\frac{1}{-\infty}=\frac{\mu-1}{\mathrm{R}} \\
\mathrm{v}_1=\frac{\mu \mathrm{R}}{\mu-1}
\end{aligned}\)
This image gets reflected from the mirror to form an image at distance \(\left\{\frac{\mu \mathrm{R}}{\mu-1}\right\}\) to the left of lens.
Refraction again from convex surface-
\(\begin{aligned}
\frac{1}{V_2}-\frac{\mu}{\frac{\mu R}{\mu-1}}=\frac{1-\mu}{-R} \\
\frac{1}{V_2}=\frac{2(\mu-1)}{R} \\
\frac{1}{f}=\frac{2(\mu-1)}{R} \\
f=\frac{R}{2(\mu-1)} \\
R^{\prime}=2 f=\frac{R}{\mu-1}
\end{aligned}\)
(left of side)
Hence, \(R^{\prime}=\frac{R}{\mu-1}\)
JIPMER-2017
Ray Optics
282353
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is \(30 \mathrm{~cm}\). The focal length of the objective, when the angular magnification of the telescope is 2 , will be:
1 \(20 \mathrm{~cm}\)
2 \(30 \mathrm{~cm}\)
3 \(10 \mathrm{~cm}\)
4 \(15 \mathrm{~cm}\)
Explanation:
AGiven, \(\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=30 \mathrm{~cm}\)
And,
\(\begin{aligned}
\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=2 \\
\mathrm{f}_0=2 \mathrm{f}_{\mathrm{e}}
\end{aligned}\)
Putting the value \(\mathrm{f}_0\) in equation (i), we get-
\(\begin{aligned}
2 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=30 \\
\mathrm{f}_{\mathrm{e}}=\frac{30}{3}=10 \mathrm{~cm}
\end{aligned}\)
Then,
\(\mathrm{f}_0=2 \mathrm{f}_{\mathrm{c}}=2 \times 10\)
So, \(\quad \mathrm{f}_0=20 \mathrm{~cm}\)
JEE Main-28.07.2022
Ray Optics
282354
The power of a lens (biconvex) is \(1.25 \mathrm{~m}^{-1}\) in particular medium. Refractive index of the lens is \(\mathbf{1 . 5}\) and radii of curvature are \(20 \mathrm{~cm}\) and 40 cm respectively. The refractive index of surrounding medium: