282487
An equiconvex (biconvex) lens has focus length f. It is cut into three parts as shown in the figure. What is the focal length of Cut part I?
1 \(3 \mathrm{f}\)
2 \(\frac{\mathrm{f}}{3}\)
3 \(\frac{\mathrm{f}}{2}\)
4 \(2 \mathrm{f}\)
Explanation:
D: If the focal length is cut into two parts in symmetrical convex lens then focal length of each part remains unchanged 'f'.
If these two parts are joined with curved ends on focal length of combination is \(\mathrm{f} / 2\). But joining two parts in opposite,
Net focal length becomes. ' \(\infty\) '.
Using lens maker formula -
\(\begin{gathered}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right) \\
\frac{1}{\mathrm{f}_1}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{gathered}\)
From equation (i) and (ii), we get-
\(\mathrm{f}_1=2 \mathrm{f} \text {. }\)
UPSEE - 2016
Ray Optics
282488
A narrow white light beam fails to converge at a point after going through a converging lens. This defect is known as
1 chromatic aberration
2 diffraction
3 polarization
4 spherical aberration
Explanation:
A: Chromatic aberration is an effect resulting from dispersion in which there is a failure of a lens to focus all colors to the same convergence point. It occurs because of the lenses have different refractive indices for different wavelength of light.
UPSEE - 2016
Ray Optics
282489
If the focal length of a lens of a camera is \(5 \mathrm{f}\) and that of another is \(2.5 \mathrm{f}\), what is the time of exposure for the second if for the first one is \(\frac{1}{200} \mathrm{~s}\) ? (where, \(\mathrm{f}\) is focal length/unit aperture)
1 \(\frac{1}{200}\)
2 \(\frac{1}{800}\)
3 \(\frac{1}{6400}\)
4 \(\frac{1}{3200}\)
Explanation:
B: Given that, Time of exposure \(\left(\mathrm{t}_1\right)=\frac{1}{200}, \mathrm{~A}_1=\mathrm{C}\left(\frac{1}{5}\right)^2, \mathrm{~A}_2=\mathrm{C}\left(\frac{1}{2.5}\right)^2\) We know that, Aperture \((\mathrm{A}) \propto\left(\frac{1}{\mathrm{f}-\text { number }}\right)^2\) Time of exposure ( \(t\) ) \(=\mathrm{C}\left(\frac{1}{\text { Aperture }}\right)^2 \ldots\) (i) \(\{\mathrm{c}\) is a constant \(\}\)
Time of exposure \((t) \propto\left(\frac{1}{\text { Aperture }}\right)^2\)
Then we have,
\(\begin{aligned}
\frac{t_1}{t_2}=\frac{\left(A_2\right)^2}{\left(A_1\right)^2} \\
t_2=\frac{t_1 A_1^2}{A_2^2} \\
t_2=\frac{1}{200}\left(\frac{2.5}{5}\right)^2 \\
t_2=\frac{1}{800} \text { sec }
\end{aligned}\)
UPSEE - 2014
Ray Optics
282499
A combination of two thin convex lenses of focal length \(0.3 \mathrm{~m}\) and \(0.1 \mathrm{~m}\) will have minimum spherical and chromatic aberrations if the distance between them is:
1 \(0.1 \mathrm{~m}\)
2 \(0.2 \mathrm{~m}\)
3 \(0.3 \mathrm{~m}\)
4 \(0.4 \mathrm{~m}\)
Explanation:
B: Given, \(\mathrm{f}_1=0.3 \mathrm{~m}, \mathrm{f}_2=0.1 \mathrm{~m}\) For minimum spherical and chromatic aberration distance between lenses.
\(\begin{aligned}
\mathrm{d}=\mathrm{f}_1-\mathrm{f}_2=0.3-0.1 \\
\mathrm{~d}=0.2 \mathrm{~m}
\end{aligned}\)
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Ray Optics
282487
An equiconvex (biconvex) lens has focus length f. It is cut into three parts as shown in the figure. What is the focal length of Cut part I?
1 \(3 \mathrm{f}\)
2 \(\frac{\mathrm{f}}{3}\)
3 \(\frac{\mathrm{f}}{2}\)
4 \(2 \mathrm{f}\)
Explanation:
D: If the focal length is cut into two parts in symmetrical convex lens then focal length of each part remains unchanged 'f'.
If these two parts are joined with curved ends on focal length of combination is \(\mathrm{f} / 2\). But joining two parts in opposite,
Net focal length becomes. ' \(\infty\) '.
Using lens maker formula -
\(\begin{gathered}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right) \\
\frac{1}{\mathrm{f}_1}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{gathered}\)
From equation (i) and (ii), we get-
\(\mathrm{f}_1=2 \mathrm{f} \text {. }\)
UPSEE - 2016
Ray Optics
282488
A narrow white light beam fails to converge at a point after going through a converging lens. This defect is known as
1 chromatic aberration
2 diffraction
3 polarization
4 spherical aberration
Explanation:
A: Chromatic aberration is an effect resulting from dispersion in which there is a failure of a lens to focus all colors to the same convergence point. It occurs because of the lenses have different refractive indices for different wavelength of light.
UPSEE - 2016
Ray Optics
282489
If the focal length of a lens of a camera is \(5 \mathrm{f}\) and that of another is \(2.5 \mathrm{f}\), what is the time of exposure for the second if for the first one is \(\frac{1}{200} \mathrm{~s}\) ? (where, \(\mathrm{f}\) is focal length/unit aperture)
1 \(\frac{1}{200}\)
2 \(\frac{1}{800}\)
3 \(\frac{1}{6400}\)
4 \(\frac{1}{3200}\)
Explanation:
B: Given that, Time of exposure \(\left(\mathrm{t}_1\right)=\frac{1}{200}, \mathrm{~A}_1=\mathrm{C}\left(\frac{1}{5}\right)^2, \mathrm{~A}_2=\mathrm{C}\left(\frac{1}{2.5}\right)^2\) We know that, Aperture \((\mathrm{A}) \propto\left(\frac{1}{\mathrm{f}-\text { number }}\right)^2\) Time of exposure ( \(t\) ) \(=\mathrm{C}\left(\frac{1}{\text { Aperture }}\right)^2 \ldots\) (i) \(\{\mathrm{c}\) is a constant \(\}\)
Time of exposure \((t) \propto\left(\frac{1}{\text { Aperture }}\right)^2\)
Then we have,
\(\begin{aligned}
\frac{t_1}{t_2}=\frac{\left(A_2\right)^2}{\left(A_1\right)^2} \\
t_2=\frac{t_1 A_1^2}{A_2^2} \\
t_2=\frac{1}{200}\left(\frac{2.5}{5}\right)^2 \\
t_2=\frac{1}{800} \text { sec }
\end{aligned}\)
UPSEE - 2014
Ray Optics
282499
A combination of two thin convex lenses of focal length \(0.3 \mathrm{~m}\) and \(0.1 \mathrm{~m}\) will have minimum spherical and chromatic aberrations if the distance between them is:
1 \(0.1 \mathrm{~m}\)
2 \(0.2 \mathrm{~m}\)
3 \(0.3 \mathrm{~m}\)
4 \(0.4 \mathrm{~m}\)
Explanation:
B: Given, \(\mathrm{f}_1=0.3 \mathrm{~m}, \mathrm{f}_2=0.1 \mathrm{~m}\) For minimum spherical and chromatic aberration distance between lenses.
\(\begin{aligned}
\mathrm{d}=\mathrm{f}_1-\mathrm{f}_2=0.3-0.1 \\
\mathrm{~d}=0.2 \mathrm{~m}
\end{aligned}\)
282487
An equiconvex (biconvex) lens has focus length f. It is cut into three parts as shown in the figure. What is the focal length of Cut part I?
1 \(3 \mathrm{f}\)
2 \(\frac{\mathrm{f}}{3}\)
3 \(\frac{\mathrm{f}}{2}\)
4 \(2 \mathrm{f}\)
Explanation:
D: If the focal length is cut into two parts in symmetrical convex lens then focal length of each part remains unchanged 'f'.
If these two parts are joined with curved ends on focal length of combination is \(\mathrm{f} / 2\). But joining two parts in opposite,
Net focal length becomes. ' \(\infty\) '.
Using lens maker formula -
\(\begin{gathered}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right) \\
\frac{1}{\mathrm{f}_1}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{gathered}\)
From equation (i) and (ii), we get-
\(\mathrm{f}_1=2 \mathrm{f} \text {. }\)
UPSEE - 2016
Ray Optics
282488
A narrow white light beam fails to converge at a point after going through a converging lens. This defect is known as
1 chromatic aberration
2 diffraction
3 polarization
4 spherical aberration
Explanation:
A: Chromatic aberration is an effect resulting from dispersion in which there is a failure of a lens to focus all colors to the same convergence point. It occurs because of the lenses have different refractive indices for different wavelength of light.
UPSEE - 2016
Ray Optics
282489
If the focal length of a lens of a camera is \(5 \mathrm{f}\) and that of another is \(2.5 \mathrm{f}\), what is the time of exposure for the second if for the first one is \(\frac{1}{200} \mathrm{~s}\) ? (where, \(\mathrm{f}\) is focal length/unit aperture)
1 \(\frac{1}{200}\)
2 \(\frac{1}{800}\)
3 \(\frac{1}{6400}\)
4 \(\frac{1}{3200}\)
Explanation:
B: Given that, Time of exposure \(\left(\mathrm{t}_1\right)=\frac{1}{200}, \mathrm{~A}_1=\mathrm{C}\left(\frac{1}{5}\right)^2, \mathrm{~A}_2=\mathrm{C}\left(\frac{1}{2.5}\right)^2\) We know that, Aperture \((\mathrm{A}) \propto\left(\frac{1}{\mathrm{f}-\text { number }}\right)^2\) Time of exposure ( \(t\) ) \(=\mathrm{C}\left(\frac{1}{\text { Aperture }}\right)^2 \ldots\) (i) \(\{\mathrm{c}\) is a constant \(\}\)
Time of exposure \((t) \propto\left(\frac{1}{\text { Aperture }}\right)^2\)
Then we have,
\(\begin{aligned}
\frac{t_1}{t_2}=\frac{\left(A_2\right)^2}{\left(A_1\right)^2} \\
t_2=\frac{t_1 A_1^2}{A_2^2} \\
t_2=\frac{1}{200}\left(\frac{2.5}{5}\right)^2 \\
t_2=\frac{1}{800} \text { sec }
\end{aligned}\)
UPSEE - 2014
Ray Optics
282499
A combination of two thin convex lenses of focal length \(0.3 \mathrm{~m}\) and \(0.1 \mathrm{~m}\) will have minimum spherical and chromatic aberrations if the distance between them is:
1 \(0.1 \mathrm{~m}\)
2 \(0.2 \mathrm{~m}\)
3 \(0.3 \mathrm{~m}\)
4 \(0.4 \mathrm{~m}\)
Explanation:
B: Given, \(\mathrm{f}_1=0.3 \mathrm{~m}, \mathrm{f}_2=0.1 \mathrm{~m}\) For minimum spherical and chromatic aberration distance between lenses.
\(\begin{aligned}
\mathrm{d}=\mathrm{f}_1-\mathrm{f}_2=0.3-0.1 \\
\mathrm{~d}=0.2 \mathrm{~m}
\end{aligned}\)
282487
An equiconvex (biconvex) lens has focus length f. It is cut into three parts as shown in the figure. What is the focal length of Cut part I?
1 \(3 \mathrm{f}\)
2 \(\frac{\mathrm{f}}{3}\)
3 \(\frac{\mathrm{f}}{2}\)
4 \(2 \mathrm{f}\)
Explanation:
D: If the focal length is cut into two parts in symmetrical convex lens then focal length of each part remains unchanged 'f'.
If these two parts are joined with curved ends on focal length of combination is \(\mathrm{f} / 2\). But joining two parts in opposite,
Net focal length becomes. ' \(\infty\) '.
Using lens maker formula -
\(\begin{gathered}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right) \\
\frac{1}{\mathrm{f}_1}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{gathered}\)
From equation (i) and (ii), we get-
\(\mathrm{f}_1=2 \mathrm{f} \text {. }\)
UPSEE - 2016
Ray Optics
282488
A narrow white light beam fails to converge at a point after going through a converging lens. This defect is known as
1 chromatic aberration
2 diffraction
3 polarization
4 spherical aberration
Explanation:
A: Chromatic aberration is an effect resulting from dispersion in which there is a failure of a lens to focus all colors to the same convergence point. It occurs because of the lenses have different refractive indices for different wavelength of light.
UPSEE - 2016
Ray Optics
282489
If the focal length of a lens of a camera is \(5 \mathrm{f}\) and that of another is \(2.5 \mathrm{f}\), what is the time of exposure for the second if for the first one is \(\frac{1}{200} \mathrm{~s}\) ? (where, \(\mathrm{f}\) is focal length/unit aperture)
1 \(\frac{1}{200}\)
2 \(\frac{1}{800}\)
3 \(\frac{1}{6400}\)
4 \(\frac{1}{3200}\)
Explanation:
B: Given that, Time of exposure \(\left(\mathrm{t}_1\right)=\frac{1}{200}, \mathrm{~A}_1=\mathrm{C}\left(\frac{1}{5}\right)^2, \mathrm{~A}_2=\mathrm{C}\left(\frac{1}{2.5}\right)^2\) We know that, Aperture \((\mathrm{A}) \propto\left(\frac{1}{\mathrm{f}-\text { number }}\right)^2\) Time of exposure ( \(t\) ) \(=\mathrm{C}\left(\frac{1}{\text { Aperture }}\right)^2 \ldots\) (i) \(\{\mathrm{c}\) is a constant \(\}\)
Time of exposure \((t) \propto\left(\frac{1}{\text { Aperture }}\right)^2\)
Then we have,
\(\begin{aligned}
\frac{t_1}{t_2}=\frac{\left(A_2\right)^2}{\left(A_1\right)^2} \\
t_2=\frac{t_1 A_1^2}{A_2^2} \\
t_2=\frac{1}{200}\left(\frac{2.5}{5}\right)^2 \\
t_2=\frac{1}{800} \text { sec }
\end{aligned}\)
UPSEE - 2014
Ray Optics
282499
A combination of two thin convex lenses of focal length \(0.3 \mathrm{~m}\) and \(0.1 \mathrm{~m}\) will have minimum spherical and chromatic aberrations if the distance between them is:
1 \(0.1 \mathrm{~m}\)
2 \(0.2 \mathrm{~m}\)
3 \(0.3 \mathrm{~m}\)
4 \(0.4 \mathrm{~m}\)
Explanation:
B: Given, \(\mathrm{f}_1=0.3 \mathrm{~m}, \mathrm{f}_2=0.1 \mathrm{~m}\) For minimum spherical and chromatic aberration distance between lenses.
\(\begin{aligned}
\mathrm{d}=\mathrm{f}_1-\mathrm{f}_2=0.3-0.1 \\
\mathrm{~d}=0.2 \mathrm{~m}
\end{aligned}\)
