282532
A thin glass (refractive index 1.5) lens has optical power of -5D in air. Its optical power in a liquid medium with refractive index 1.6 will be
1 \(-1 \mathrm{D}\)
2 \(1 \mathrm{D}\)
3 \(-25 \mathrm{D}\)
4 \(25 \mathrm{D}\)
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Power \((\mathrm{P})=-5 \mathrm{D}\)
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.6\)
\(\therefore \quad\) Power, \(P=\frac{100}{\mathrm{f}}\)
\(\mathrm{f}=\frac{100}{-5}=-20 \mathrm{~cm}\)
Focal length in liquid -
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mu_{\mathrm{g}}-1}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}\right)-1}\right) \\
\left.\mathrm{f}^{\prime}=(-20)\left(\frac{1.5-1}{\left(\frac{1.5}{1.6}-1\right)}\right)\right) \\
\mathrm{f}^{\prime}=20 \times 8 \\
\mathrm{f}^{\prime}=160 \mathrm{~cm}=1.6 \mathrm{~m}
\end{aligned}\)
Hence, power, \(\mathrm{P}^{\prime}=\frac{\mu_{\mathrm{L}}}{\mathrm{f}^{\prime}}\)
\(\begin{gathered}
\mathrm{P}^{\prime}=\frac{1.6}{1.6} \\
\mathrm{P}^{\prime}=1 \mathrm{D}
\end{gathered}\)
AIIMS-2008
Ray Optics
282554
A focal length of a lens is \(10 \mathrm{~cm}\). What is power of a lens in dipotre?
1 \(0.1 \mathrm{D}\)
2 \(10 \mathrm{D}\)
3 \(15 \mathrm{D}\)
4 \(1 \mathrm{D}\)
Explanation:
B: Given, \(\mathrm{f}=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Power of lens, \(P=\frac{1}{f}=\frac{1}{0.1}=10 \mathrm{D}\)
\(\mathrm{P}=10 \mathrm{D}\)
Karnataka CET-2014
Ray Optics
282533
The focal length of the objective and eye lenses of a microscope are \(1.6 \mathrm{~cm}\) and \(2.5 \mathrm{~cm}\) respectively. The distance between the two lenses is \(21.7 \mathrm{~cm}\). If the final image is formed at infinity, what is the linear magnification?
1 11
2 110
3 1.1
4 44
Explanation:
B Given,
Focal length of objective lens \(\left(f_0\right)=1.6 \mathrm{~cm}\)
Focal length of eye piece lens \(\left(f_e\right)=2.5 \mathrm{~cm}\)
\(\mathrm{d}=21.7 \mathrm{~cm}, \mathrm{v}=\infty\)
We know that,
\(\text { Magnification }(m)=\frac{\text { L.d }}{f_{\mathrm{e}} \mathrm{f}_{\mathrm{o}}}\)
Here, \(L=d-f_o=21.7-1.6=20.1\)
\(\begin{aligned}
\therefore \quad & \mathrm{m}=\frac{20.1 \times 21.7}{2.5 \times 1.6} \\
\mathrm{~m}=109.1 \\
\mathrm{~m} \sqcup 110
\end{aligned}\)
Hence, linear magnification of microscope \((\mathrm{m})=110\)
AIIMS-2007
Ray Optics
282534
A lens is made of flint glass (refractive index \(=\) 1.5). When the lens is immersed in a liquid of refractive index 1.25 , the focal length:
1 increases by a factor of 1.25
2 increases by a factor of 2.5
3 increases by a factor of 1.2
4 decreases by a factor of 1.2
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.25\)
New focal length is-
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f} \frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}-1\right)}=\mathrm{f} \frac{(1.5-1)}{\left(\frac{1.5}{1.25}-1\right)} \\
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{0.5}{0.2}\right) \\
\mathrm{f}^{\prime}=2.5 \mathrm{f}
\end{aligned}\)
282532
A thin glass (refractive index 1.5) lens has optical power of -5D in air. Its optical power in a liquid medium with refractive index 1.6 will be
1 \(-1 \mathrm{D}\)
2 \(1 \mathrm{D}\)
3 \(-25 \mathrm{D}\)
4 \(25 \mathrm{D}\)
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Power \((\mathrm{P})=-5 \mathrm{D}\)
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.6\)
\(\therefore \quad\) Power, \(P=\frac{100}{\mathrm{f}}\)
\(\mathrm{f}=\frac{100}{-5}=-20 \mathrm{~cm}\)
Focal length in liquid -
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mu_{\mathrm{g}}-1}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}\right)-1}\right) \\
\left.\mathrm{f}^{\prime}=(-20)\left(\frac{1.5-1}{\left(\frac{1.5}{1.6}-1\right)}\right)\right) \\
\mathrm{f}^{\prime}=20 \times 8 \\
\mathrm{f}^{\prime}=160 \mathrm{~cm}=1.6 \mathrm{~m}
\end{aligned}\)
Hence, power, \(\mathrm{P}^{\prime}=\frac{\mu_{\mathrm{L}}}{\mathrm{f}^{\prime}}\)
\(\begin{gathered}
\mathrm{P}^{\prime}=\frac{1.6}{1.6} \\
\mathrm{P}^{\prime}=1 \mathrm{D}
\end{gathered}\)
AIIMS-2008
Ray Optics
282554
A focal length of a lens is \(10 \mathrm{~cm}\). What is power of a lens in dipotre?
1 \(0.1 \mathrm{D}\)
2 \(10 \mathrm{D}\)
3 \(15 \mathrm{D}\)
4 \(1 \mathrm{D}\)
Explanation:
B: Given, \(\mathrm{f}=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Power of lens, \(P=\frac{1}{f}=\frac{1}{0.1}=10 \mathrm{D}\)
\(\mathrm{P}=10 \mathrm{D}\)
Karnataka CET-2014
Ray Optics
282533
The focal length of the objective and eye lenses of a microscope are \(1.6 \mathrm{~cm}\) and \(2.5 \mathrm{~cm}\) respectively. The distance between the two lenses is \(21.7 \mathrm{~cm}\). If the final image is formed at infinity, what is the linear magnification?
1 11
2 110
3 1.1
4 44
Explanation:
B Given,
Focal length of objective lens \(\left(f_0\right)=1.6 \mathrm{~cm}\)
Focal length of eye piece lens \(\left(f_e\right)=2.5 \mathrm{~cm}\)
\(\mathrm{d}=21.7 \mathrm{~cm}, \mathrm{v}=\infty\)
We know that,
\(\text { Magnification }(m)=\frac{\text { L.d }}{f_{\mathrm{e}} \mathrm{f}_{\mathrm{o}}}\)
Here, \(L=d-f_o=21.7-1.6=20.1\)
\(\begin{aligned}
\therefore \quad & \mathrm{m}=\frac{20.1 \times 21.7}{2.5 \times 1.6} \\
\mathrm{~m}=109.1 \\
\mathrm{~m} \sqcup 110
\end{aligned}\)
Hence, linear magnification of microscope \((\mathrm{m})=110\)
AIIMS-2007
Ray Optics
282534
A lens is made of flint glass (refractive index \(=\) 1.5). When the lens is immersed in a liquid of refractive index 1.25 , the focal length:
1 increases by a factor of 1.25
2 increases by a factor of 2.5
3 increases by a factor of 1.2
4 decreases by a factor of 1.2
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.25\)
New focal length is-
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f} \frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}-1\right)}=\mathrm{f} \frac{(1.5-1)}{\left(\frac{1.5}{1.25}-1\right)} \\
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{0.5}{0.2}\right) \\
\mathrm{f}^{\prime}=2.5 \mathrm{f}
\end{aligned}\)
282532
A thin glass (refractive index 1.5) lens has optical power of -5D in air. Its optical power in a liquid medium with refractive index 1.6 will be
1 \(-1 \mathrm{D}\)
2 \(1 \mathrm{D}\)
3 \(-25 \mathrm{D}\)
4 \(25 \mathrm{D}\)
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Power \((\mathrm{P})=-5 \mathrm{D}\)
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.6\)
\(\therefore \quad\) Power, \(P=\frac{100}{\mathrm{f}}\)
\(\mathrm{f}=\frac{100}{-5}=-20 \mathrm{~cm}\)
Focal length in liquid -
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mu_{\mathrm{g}}-1}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}\right)-1}\right) \\
\left.\mathrm{f}^{\prime}=(-20)\left(\frac{1.5-1}{\left(\frac{1.5}{1.6}-1\right)}\right)\right) \\
\mathrm{f}^{\prime}=20 \times 8 \\
\mathrm{f}^{\prime}=160 \mathrm{~cm}=1.6 \mathrm{~m}
\end{aligned}\)
Hence, power, \(\mathrm{P}^{\prime}=\frac{\mu_{\mathrm{L}}}{\mathrm{f}^{\prime}}\)
\(\begin{gathered}
\mathrm{P}^{\prime}=\frac{1.6}{1.6} \\
\mathrm{P}^{\prime}=1 \mathrm{D}
\end{gathered}\)
AIIMS-2008
Ray Optics
282554
A focal length of a lens is \(10 \mathrm{~cm}\). What is power of a lens in dipotre?
1 \(0.1 \mathrm{D}\)
2 \(10 \mathrm{D}\)
3 \(15 \mathrm{D}\)
4 \(1 \mathrm{D}\)
Explanation:
B: Given, \(\mathrm{f}=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Power of lens, \(P=\frac{1}{f}=\frac{1}{0.1}=10 \mathrm{D}\)
\(\mathrm{P}=10 \mathrm{D}\)
Karnataka CET-2014
Ray Optics
282533
The focal length of the objective and eye lenses of a microscope are \(1.6 \mathrm{~cm}\) and \(2.5 \mathrm{~cm}\) respectively. The distance between the two lenses is \(21.7 \mathrm{~cm}\). If the final image is formed at infinity, what is the linear magnification?
1 11
2 110
3 1.1
4 44
Explanation:
B Given,
Focal length of objective lens \(\left(f_0\right)=1.6 \mathrm{~cm}\)
Focal length of eye piece lens \(\left(f_e\right)=2.5 \mathrm{~cm}\)
\(\mathrm{d}=21.7 \mathrm{~cm}, \mathrm{v}=\infty\)
We know that,
\(\text { Magnification }(m)=\frac{\text { L.d }}{f_{\mathrm{e}} \mathrm{f}_{\mathrm{o}}}\)
Here, \(L=d-f_o=21.7-1.6=20.1\)
\(\begin{aligned}
\therefore \quad & \mathrm{m}=\frac{20.1 \times 21.7}{2.5 \times 1.6} \\
\mathrm{~m}=109.1 \\
\mathrm{~m} \sqcup 110
\end{aligned}\)
Hence, linear magnification of microscope \((\mathrm{m})=110\)
AIIMS-2007
Ray Optics
282534
A lens is made of flint glass (refractive index \(=\) 1.5). When the lens is immersed in a liquid of refractive index 1.25 , the focal length:
1 increases by a factor of 1.25
2 increases by a factor of 2.5
3 increases by a factor of 1.2
4 decreases by a factor of 1.2
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.25\)
New focal length is-
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f} \frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}-1\right)}=\mathrm{f} \frac{(1.5-1)}{\left(\frac{1.5}{1.25}-1\right)} \\
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{0.5}{0.2}\right) \\
\mathrm{f}^{\prime}=2.5 \mathrm{f}
\end{aligned}\)
282532
A thin glass (refractive index 1.5) lens has optical power of -5D in air. Its optical power in a liquid medium with refractive index 1.6 will be
1 \(-1 \mathrm{D}\)
2 \(1 \mathrm{D}\)
3 \(-25 \mathrm{D}\)
4 \(25 \mathrm{D}\)
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Power \((\mathrm{P})=-5 \mathrm{D}\)
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.6\)
\(\therefore \quad\) Power, \(P=\frac{100}{\mathrm{f}}\)
\(\mathrm{f}=\frac{100}{-5}=-20 \mathrm{~cm}\)
Focal length in liquid -
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mu_{\mathrm{g}}-1}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}\right)-1}\right) \\
\left.\mathrm{f}^{\prime}=(-20)\left(\frac{1.5-1}{\left(\frac{1.5}{1.6}-1\right)}\right)\right) \\
\mathrm{f}^{\prime}=20 \times 8 \\
\mathrm{f}^{\prime}=160 \mathrm{~cm}=1.6 \mathrm{~m}
\end{aligned}\)
Hence, power, \(\mathrm{P}^{\prime}=\frac{\mu_{\mathrm{L}}}{\mathrm{f}^{\prime}}\)
\(\begin{gathered}
\mathrm{P}^{\prime}=\frac{1.6}{1.6} \\
\mathrm{P}^{\prime}=1 \mathrm{D}
\end{gathered}\)
AIIMS-2008
Ray Optics
282554
A focal length of a lens is \(10 \mathrm{~cm}\). What is power of a lens in dipotre?
1 \(0.1 \mathrm{D}\)
2 \(10 \mathrm{D}\)
3 \(15 \mathrm{D}\)
4 \(1 \mathrm{D}\)
Explanation:
B: Given, \(\mathrm{f}=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Power of lens, \(P=\frac{1}{f}=\frac{1}{0.1}=10 \mathrm{D}\)
\(\mathrm{P}=10 \mathrm{D}\)
Karnataka CET-2014
Ray Optics
282533
The focal length of the objective and eye lenses of a microscope are \(1.6 \mathrm{~cm}\) and \(2.5 \mathrm{~cm}\) respectively. The distance between the two lenses is \(21.7 \mathrm{~cm}\). If the final image is formed at infinity, what is the linear magnification?
1 11
2 110
3 1.1
4 44
Explanation:
B Given,
Focal length of objective lens \(\left(f_0\right)=1.6 \mathrm{~cm}\)
Focal length of eye piece lens \(\left(f_e\right)=2.5 \mathrm{~cm}\)
\(\mathrm{d}=21.7 \mathrm{~cm}, \mathrm{v}=\infty\)
We know that,
\(\text { Magnification }(m)=\frac{\text { L.d }}{f_{\mathrm{e}} \mathrm{f}_{\mathrm{o}}}\)
Here, \(L=d-f_o=21.7-1.6=20.1\)
\(\begin{aligned}
\therefore \quad & \mathrm{m}=\frac{20.1 \times 21.7}{2.5 \times 1.6} \\
\mathrm{~m}=109.1 \\
\mathrm{~m} \sqcup 110
\end{aligned}\)
Hence, linear magnification of microscope \((\mathrm{m})=110\)
AIIMS-2007
Ray Optics
282534
A lens is made of flint glass (refractive index \(=\) 1.5). When the lens is immersed in a liquid of refractive index 1.25 , the focal length:
1 increases by a factor of 1.25
2 increases by a factor of 2.5
3 increases by a factor of 1.2
4 decreases by a factor of 1.2
Explanation:
BGiven,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
Refractive index of liquid \(\left(\mu_{\mathrm{L}}\right)=1.25\)
New focal length is-
\(\begin{aligned}
\mathrm{f}^{\prime}=\mathrm{f} \frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\mu_{\mathrm{g}} / \mu_{\mathrm{L}}-1\right)}=\mathrm{f} \frac{(1.5-1)}{\left(\frac{1.5}{1.25}-1\right)} \\
\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{0.5}{0.2}\right) \\
\mathrm{f}^{\prime}=2.5 \mathrm{f}
\end{aligned}\)
