282371
The power of a biconvex lens is \(10 \mathrm{D}\) and the radius of curvature of each surface is \(10 \mathrm{~cm}\). Then, the refractive index of the material of the lens is
282372
The power of two lenses are \(2.5 \mathrm{D}\) and \(1.5 \mathrm{D}\) respectively. If they kept in proximity, the power of the combination will be
1 \(1.0 \mathrm{D}\)
2 \(\frac{5}{3} \mathrm{D}\)
3 \(\frac{5}{3} \mathrm{D}\)
4 \(4.0 \mathrm{D}\)
Explanation:
D: Given, \(\mathrm{P}_1=2.5 \mathrm{D} \& \mathrm{P}_2=1.5 \mathrm{D}\)
Combined power is given by -
\(\begin{aligned}
\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2 \\
\mathrm{P}=2.5+1.5 \\
\mathrm{P}=4 \mathrm{D}
\end{aligned}\)
Tripura-2020
Ray Optics
282373
A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature \(R\). On immersion in a medium of refractive index 1.75 , it will behave as a
1 Convergent lens of focal length \(3.5 \mathrm{R}\)
2 Convergent lens of focal length \(3.0 \mathrm{R}\)
3 Divergent lens of focal length \(3.5 \mathrm{R}\)
4 Divergent lens of focal length \(3.0 \mathrm{R}\)
Explanation:
A: Given, \(\mu_{\mathrm{g}}=1.5\)
\(\mu_{\mathrm{L}}=1.75\)
We know that, Applying lens formula between glass and air-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midum }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{\mu_{\text {air }}}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{1}-1\right)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=-\frac{1}{\mathrm{R}} \\
\therefore \quad \mathrm{f}=-\mathrm{R} \\
&
\end{aligned}\)
Now, applying lens maker formula between glass of refractive index of 1.5 and refractive index 1.75
\(\begin{aligned}
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midium }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.75}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{-0.25}{1.75}\left(\frac{-2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{1}{7} \times \frac{2}{\mathrm{R}} \\
\mathrm{f}^{\prime}=\frac{7}{2} \mathrm{R} \\
\mathrm{f}^{\prime}=3.5 \mathrm{R}
\end{aligned}\)
\(\therefore\) The lens behave as convex or convergent lens with focal length of \(3.5 \mathrm{R}\).
AP EAMCET-07.10.2020
Ray Optics
282374
A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length \(\frac{5}{3}\). If their equivalent focal length is \(45 \mathrm{~cm}\). the individual focal lengths of concave and convex lens are respectively.
1 \(-10 \mathrm{~cm} .15 \mathrm{~cm}\)
2 \(-25 \mathrm{~cm} .10 \mathrm{~cm}\)
3 \(-85 \mathrm{~cm} .10 \mathrm{~cm}\)
4 \(-30 \mathrm{~cm} .18 \mathrm{~cm}\)
Explanation:
D: Let focal length of convex lens \(=\mathrm{f}_1\)
Focal length of concave lens \(=f_2\)
\(\frac{\left|\mathrm{f}_1\right|}{\left|\mathrm{f}_2\right|}=\frac{5}{3}\)
Combined focal length-
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{45}=\frac{1}{\mathrm{f}_1}-\frac{5}{3 \mathrm{f}_1} \\
\frac{1}{45}=\frac{3-5}{3 \mathrm{f}_1}
\end{aligned}\)
\(f_1=-30 \mathrm{~cm}\{\) Negative sign because of concave lens \(\}\)
And,
\(f_2=-30 \times\left(\frac{-3}{5}\right)\)
\(\mathrm{f}_2=18 \mathrm{~cm}\)
282371
The power of a biconvex lens is \(10 \mathrm{D}\) and the radius of curvature of each surface is \(10 \mathrm{~cm}\). Then, the refractive index of the material of the lens is
282372
The power of two lenses are \(2.5 \mathrm{D}\) and \(1.5 \mathrm{D}\) respectively. If they kept in proximity, the power of the combination will be
1 \(1.0 \mathrm{D}\)
2 \(\frac{5}{3} \mathrm{D}\)
3 \(\frac{5}{3} \mathrm{D}\)
4 \(4.0 \mathrm{D}\)
Explanation:
D: Given, \(\mathrm{P}_1=2.5 \mathrm{D} \& \mathrm{P}_2=1.5 \mathrm{D}\)
Combined power is given by -
\(\begin{aligned}
\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2 \\
\mathrm{P}=2.5+1.5 \\
\mathrm{P}=4 \mathrm{D}
\end{aligned}\)
Tripura-2020
Ray Optics
282373
A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature \(R\). On immersion in a medium of refractive index 1.75 , it will behave as a
1 Convergent lens of focal length \(3.5 \mathrm{R}\)
2 Convergent lens of focal length \(3.0 \mathrm{R}\)
3 Divergent lens of focal length \(3.5 \mathrm{R}\)
4 Divergent lens of focal length \(3.0 \mathrm{R}\)
Explanation:
A: Given, \(\mu_{\mathrm{g}}=1.5\)
\(\mu_{\mathrm{L}}=1.75\)
We know that, Applying lens formula between glass and air-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midum }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{\mu_{\text {air }}}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{1}-1\right)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=-\frac{1}{\mathrm{R}} \\
\therefore \quad \mathrm{f}=-\mathrm{R} \\
&
\end{aligned}\)
Now, applying lens maker formula between glass of refractive index of 1.5 and refractive index 1.75
\(\begin{aligned}
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midium }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.75}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{-0.25}{1.75}\left(\frac{-2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{1}{7} \times \frac{2}{\mathrm{R}} \\
\mathrm{f}^{\prime}=\frac{7}{2} \mathrm{R} \\
\mathrm{f}^{\prime}=3.5 \mathrm{R}
\end{aligned}\)
\(\therefore\) The lens behave as convex or convergent lens with focal length of \(3.5 \mathrm{R}\).
AP EAMCET-07.10.2020
Ray Optics
282374
A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length \(\frac{5}{3}\). If their equivalent focal length is \(45 \mathrm{~cm}\). the individual focal lengths of concave and convex lens are respectively.
1 \(-10 \mathrm{~cm} .15 \mathrm{~cm}\)
2 \(-25 \mathrm{~cm} .10 \mathrm{~cm}\)
3 \(-85 \mathrm{~cm} .10 \mathrm{~cm}\)
4 \(-30 \mathrm{~cm} .18 \mathrm{~cm}\)
Explanation:
D: Let focal length of convex lens \(=\mathrm{f}_1\)
Focal length of concave lens \(=f_2\)
\(\frac{\left|\mathrm{f}_1\right|}{\left|\mathrm{f}_2\right|}=\frac{5}{3}\)
Combined focal length-
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{45}=\frac{1}{\mathrm{f}_1}-\frac{5}{3 \mathrm{f}_1} \\
\frac{1}{45}=\frac{3-5}{3 \mathrm{f}_1}
\end{aligned}\)
\(f_1=-30 \mathrm{~cm}\{\) Negative sign because of concave lens \(\}\)
And,
\(f_2=-30 \times\left(\frac{-3}{5}\right)\)
\(\mathrm{f}_2=18 \mathrm{~cm}\)
282371
The power of a biconvex lens is \(10 \mathrm{D}\) and the radius of curvature of each surface is \(10 \mathrm{~cm}\). Then, the refractive index of the material of the lens is
282372
The power of two lenses are \(2.5 \mathrm{D}\) and \(1.5 \mathrm{D}\) respectively. If they kept in proximity, the power of the combination will be
1 \(1.0 \mathrm{D}\)
2 \(\frac{5}{3} \mathrm{D}\)
3 \(\frac{5}{3} \mathrm{D}\)
4 \(4.0 \mathrm{D}\)
Explanation:
D: Given, \(\mathrm{P}_1=2.5 \mathrm{D} \& \mathrm{P}_2=1.5 \mathrm{D}\)
Combined power is given by -
\(\begin{aligned}
\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2 \\
\mathrm{P}=2.5+1.5 \\
\mathrm{P}=4 \mathrm{D}
\end{aligned}\)
Tripura-2020
Ray Optics
282373
A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature \(R\). On immersion in a medium of refractive index 1.75 , it will behave as a
1 Convergent lens of focal length \(3.5 \mathrm{R}\)
2 Convergent lens of focal length \(3.0 \mathrm{R}\)
3 Divergent lens of focal length \(3.5 \mathrm{R}\)
4 Divergent lens of focal length \(3.0 \mathrm{R}\)
Explanation:
A: Given, \(\mu_{\mathrm{g}}=1.5\)
\(\mu_{\mathrm{L}}=1.75\)
We know that, Applying lens formula between glass and air-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midum }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{\mu_{\text {air }}}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{1}-1\right)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=-\frac{1}{\mathrm{R}} \\
\therefore \quad \mathrm{f}=-\mathrm{R} \\
&
\end{aligned}\)
Now, applying lens maker formula between glass of refractive index of 1.5 and refractive index 1.75
\(\begin{aligned}
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midium }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.75}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{-0.25}{1.75}\left(\frac{-2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{1}{7} \times \frac{2}{\mathrm{R}} \\
\mathrm{f}^{\prime}=\frac{7}{2} \mathrm{R} \\
\mathrm{f}^{\prime}=3.5 \mathrm{R}
\end{aligned}\)
\(\therefore\) The lens behave as convex or convergent lens with focal length of \(3.5 \mathrm{R}\).
AP EAMCET-07.10.2020
Ray Optics
282374
A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length \(\frac{5}{3}\). If their equivalent focal length is \(45 \mathrm{~cm}\). the individual focal lengths of concave and convex lens are respectively.
1 \(-10 \mathrm{~cm} .15 \mathrm{~cm}\)
2 \(-25 \mathrm{~cm} .10 \mathrm{~cm}\)
3 \(-85 \mathrm{~cm} .10 \mathrm{~cm}\)
4 \(-30 \mathrm{~cm} .18 \mathrm{~cm}\)
Explanation:
D: Let focal length of convex lens \(=\mathrm{f}_1\)
Focal length of concave lens \(=f_2\)
\(\frac{\left|\mathrm{f}_1\right|}{\left|\mathrm{f}_2\right|}=\frac{5}{3}\)
Combined focal length-
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{45}=\frac{1}{\mathrm{f}_1}-\frac{5}{3 \mathrm{f}_1} \\
\frac{1}{45}=\frac{3-5}{3 \mathrm{f}_1}
\end{aligned}\)
\(f_1=-30 \mathrm{~cm}\{\) Negative sign because of concave lens \(\}\)
And,
\(f_2=-30 \times\left(\frac{-3}{5}\right)\)
\(\mathrm{f}_2=18 \mathrm{~cm}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ray Optics
282371
The power of a biconvex lens is \(10 \mathrm{D}\) and the radius of curvature of each surface is \(10 \mathrm{~cm}\). Then, the refractive index of the material of the lens is
282372
The power of two lenses are \(2.5 \mathrm{D}\) and \(1.5 \mathrm{D}\) respectively. If they kept in proximity, the power of the combination will be
1 \(1.0 \mathrm{D}\)
2 \(\frac{5}{3} \mathrm{D}\)
3 \(\frac{5}{3} \mathrm{D}\)
4 \(4.0 \mathrm{D}\)
Explanation:
D: Given, \(\mathrm{P}_1=2.5 \mathrm{D} \& \mathrm{P}_2=1.5 \mathrm{D}\)
Combined power is given by -
\(\begin{aligned}
\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2 \\
\mathrm{P}=2.5+1.5 \\
\mathrm{P}=4 \mathrm{D}
\end{aligned}\)
Tripura-2020
Ray Optics
282373
A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature \(R\). On immersion in a medium of refractive index 1.75 , it will behave as a
1 Convergent lens of focal length \(3.5 \mathrm{R}\)
2 Convergent lens of focal length \(3.0 \mathrm{R}\)
3 Divergent lens of focal length \(3.5 \mathrm{R}\)
4 Divergent lens of focal length \(3.0 \mathrm{R}\)
Explanation:
A: Given, \(\mu_{\mathrm{g}}=1.5\)
\(\mu_{\mathrm{L}}=1.75\)
We know that, Applying lens formula between glass and air-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midum }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{\mu_{\text {air }}}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)
\end{aligned}\)
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\frac{1.5}{1}-1\right)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=-\frac{1}{\mathrm{R}} \\
\therefore \quad \mathrm{f}=-\mathrm{R} \\
&
\end{aligned}\)
Now, applying lens maker formula between glass of refractive index of 1.5 and refractive index 1.75
\(\begin{aligned}
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {midium }}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.75}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{-0.25}{1.75}\left(\frac{-2}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=\frac{1}{7} \times \frac{2}{\mathrm{R}} \\
\mathrm{f}^{\prime}=\frac{7}{2} \mathrm{R} \\
\mathrm{f}^{\prime}=3.5 \mathrm{R}
\end{aligned}\)
\(\therefore\) The lens behave as convex or convergent lens with focal length of \(3.5 \mathrm{R}\).
AP EAMCET-07.10.2020
Ray Optics
282374
A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length \(\frac{5}{3}\). If their equivalent focal length is \(45 \mathrm{~cm}\). the individual focal lengths of concave and convex lens are respectively.
1 \(-10 \mathrm{~cm} .15 \mathrm{~cm}\)
2 \(-25 \mathrm{~cm} .10 \mathrm{~cm}\)
3 \(-85 \mathrm{~cm} .10 \mathrm{~cm}\)
4 \(-30 \mathrm{~cm} .18 \mathrm{~cm}\)
Explanation:
D: Let focal length of convex lens \(=\mathrm{f}_1\)
Focal length of concave lens \(=f_2\)
\(\frac{\left|\mathrm{f}_1\right|}{\left|\mathrm{f}_2\right|}=\frac{5}{3}\)
Combined focal length-
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{45}=\frac{1}{\mathrm{f}_1}-\frac{5}{3 \mathrm{f}_1} \\
\frac{1}{45}=\frac{3-5}{3 \mathrm{f}_1}
\end{aligned}\)
\(f_1=-30 \mathrm{~cm}\{\) Negative sign because of concave lens \(\}\)
And,
\(f_2=-30 \times\left(\frac{-3}{5}\right)\)
\(\mathrm{f}_2=18 \mathrm{~cm}\)
