282479
A person suffering from hypermetropia requires which type of spectacle lenses?
1 Concave
2 Plano- concave
3 Convex
4 Cylindrical
Explanation:
C: Hypermetropia -: When the human eye can see distant object clearly, but cannot see near by distinctly. A convex lens of suitable focal length will correct hypermetropia.
CG PET- 2005
Ray Optics
282480
The focal length of a convex lens is \(10 \mathrm{~cm}\) and its refractive index is 1.5 . If the radius of curvature of one surface in \(7.5 \mathrm{~cm}\), the radius of curvature of the second surface will be
1 \(7.5 \mathrm{~cm}\)
2 \(15.0 \mathrm{~cm}\)
3 \(75 \mathrm{~cm}\)
4 \(5.0 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of convex lens \(=10 \mathrm{~cm}\)
Refractive index \((\mu)=1.5\)
Radius of curvature of \(\mathrm{I}^{\mathrm{st}}\) surface \(\left(\mathrm{R}_1\right)=7.5 \mathrm{~cm}\)
Using power of lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{10}=(1.5-1)\left(\frac{1}{7.5}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{5}=\frac{1}{7.5}-\frac{1}{\mathrm{R}_2} \\
\frac{1}{\mathrm{R}_2}=\frac{1}{7.5}-\frac{1}{5} \\
\frac{1}{\mathrm{R}_2}=\frac{-0.5}{7.5} \\
\mathrm{R}_2=15 \mathrm{~cm}
\end{aligned}\)
CG PET- 2004
Ray Optics
282481
A plano-convex lens of focal length \(30 \mathrm{~cm}\) has its plane surface silvered. An object is placed \(40 \mathrm{~cm}\) from the lens on the convex side. The distance of the image from the lens is
1 \(18 \mathrm{~cm}\)
2 \(24 \mathrm{~cm}\)
3 \(30 \mathrm{~cm}\)
4 \(40 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of plano convex lens (f) \(=30 \mathrm{~cm}\)
Distance of object \((u)=-40 \mathrm{~cm}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{v}}=\frac{1}{30}-\frac{1}{40} \\
\frac{1}{\mathrm{v}}=\frac{4-3}{120} \\
\mathrm{v}=120 \mathrm{~cm}=\mathrm{u}^{\prime}
\end{aligned}\)
Now, this image act as object for the lens.
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{(-120)} \quad\left[\because \mathrm{u}^{\prime}=-120 \mathrm{~cm}\right] \\
\frac{1}{\mathrm{v}^{\prime}}=\frac{1}{30}+\frac{1}{120} \\
\mathrm{v}^{\prime}=24 \mathrm{~cm}
\end{aligned}\)
BITSAT-2009
Ray Optics
282482
A thin convergent glass lens \(\left(\mu_{\mathrm{g}}=1.5\right)\) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index \(\mu\), it acts as a divergent lens of focal length \(100 \mathrm{~cm}\). The value of \(\mu\) must be
1 \(4 / 3\)
2 \(5 / 3\)
3 \(5 / 4(\mathrm{~d})\)
\(6 / 5\)
Explanation:
B: Given,
Refractive index of air \(\left(\mu_{\mathrm{a}}\right)=1\)
Refractive index of thin convergent lens \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Refractive index of liquid \(=\mu\)
Focal length of divergent lens, \(f=-10 \mathrm{~cm}\)
Power of lens,
\(\mathrm{P}_{\mathrm{c}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
When lens immersed in liquid of refractive index ' \(\mu\) '. Focal length formula -
\(\mathrm{P}_{\mathrm{d}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) / (ii)
\(\frac{P_c}{P_d}=\frac{\left(\frac{\mu_g}{\mu}-1\right)}{\left(\frac{\mu_g}{\mu}-1\right)}\)
\(\frac{5}{-1}=\left(\frac{1.5-1}{\frac{1.5}{\mu}-1}\right) \quad\left\{\therefore \mathrm{P}_{\mathrm{d}}=\frac{100}{\mathrm{f}}=\frac{100}{-100}=-1\right\}\)
\(-5=\frac{0.5 \mu}{1.5-\mu}\)
\(-7.5+5 \mu=0.5 \mu\)
\(\mu=\frac{7.5}{4.5}\)
\(\mu=\frac{5}{3}\)
282479
A person suffering from hypermetropia requires which type of spectacle lenses?
1 Concave
2 Plano- concave
3 Convex
4 Cylindrical
Explanation:
C: Hypermetropia -: When the human eye can see distant object clearly, but cannot see near by distinctly. A convex lens of suitable focal length will correct hypermetropia.
CG PET- 2005
Ray Optics
282480
The focal length of a convex lens is \(10 \mathrm{~cm}\) and its refractive index is 1.5 . If the radius of curvature of one surface in \(7.5 \mathrm{~cm}\), the radius of curvature of the second surface will be
1 \(7.5 \mathrm{~cm}\)
2 \(15.0 \mathrm{~cm}\)
3 \(75 \mathrm{~cm}\)
4 \(5.0 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of convex lens \(=10 \mathrm{~cm}\)
Refractive index \((\mu)=1.5\)
Radius of curvature of \(\mathrm{I}^{\mathrm{st}}\) surface \(\left(\mathrm{R}_1\right)=7.5 \mathrm{~cm}\)
Using power of lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{10}=(1.5-1)\left(\frac{1}{7.5}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{5}=\frac{1}{7.5}-\frac{1}{\mathrm{R}_2} \\
\frac{1}{\mathrm{R}_2}=\frac{1}{7.5}-\frac{1}{5} \\
\frac{1}{\mathrm{R}_2}=\frac{-0.5}{7.5} \\
\mathrm{R}_2=15 \mathrm{~cm}
\end{aligned}\)
CG PET- 2004
Ray Optics
282481
A plano-convex lens of focal length \(30 \mathrm{~cm}\) has its plane surface silvered. An object is placed \(40 \mathrm{~cm}\) from the lens on the convex side. The distance of the image from the lens is
1 \(18 \mathrm{~cm}\)
2 \(24 \mathrm{~cm}\)
3 \(30 \mathrm{~cm}\)
4 \(40 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of plano convex lens (f) \(=30 \mathrm{~cm}\)
Distance of object \((u)=-40 \mathrm{~cm}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{v}}=\frac{1}{30}-\frac{1}{40} \\
\frac{1}{\mathrm{v}}=\frac{4-3}{120} \\
\mathrm{v}=120 \mathrm{~cm}=\mathrm{u}^{\prime}
\end{aligned}\)
Now, this image act as object for the lens.
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{(-120)} \quad\left[\because \mathrm{u}^{\prime}=-120 \mathrm{~cm}\right] \\
\frac{1}{\mathrm{v}^{\prime}}=\frac{1}{30}+\frac{1}{120} \\
\mathrm{v}^{\prime}=24 \mathrm{~cm}
\end{aligned}\)
BITSAT-2009
Ray Optics
282482
A thin convergent glass lens \(\left(\mu_{\mathrm{g}}=1.5\right)\) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index \(\mu\), it acts as a divergent lens of focal length \(100 \mathrm{~cm}\). The value of \(\mu\) must be
1 \(4 / 3\)
2 \(5 / 3\)
3 \(5 / 4(\mathrm{~d})\)
\(6 / 5\)
Explanation:
B: Given,
Refractive index of air \(\left(\mu_{\mathrm{a}}\right)=1\)
Refractive index of thin convergent lens \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Refractive index of liquid \(=\mu\)
Focal length of divergent lens, \(f=-10 \mathrm{~cm}\)
Power of lens,
\(\mathrm{P}_{\mathrm{c}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
When lens immersed in liquid of refractive index ' \(\mu\) '. Focal length formula -
\(\mathrm{P}_{\mathrm{d}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) / (ii)
\(\frac{P_c}{P_d}=\frac{\left(\frac{\mu_g}{\mu}-1\right)}{\left(\frac{\mu_g}{\mu}-1\right)}\)
\(\frac{5}{-1}=\left(\frac{1.5-1}{\frac{1.5}{\mu}-1}\right) \quad\left\{\therefore \mathrm{P}_{\mathrm{d}}=\frac{100}{\mathrm{f}}=\frac{100}{-100}=-1\right\}\)
\(-5=\frac{0.5 \mu}{1.5-\mu}\)
\(-7.5+5 \mu=0.5 \mu\)
\(\mu=\frac{7.5}{4.5}\)
\(\mu=\frac{5}{3}\)
282479
A person suffering from hypermetropia requires which type of spectacle lenses?
1 Concave
2 Plano- concave
3 Convex
4 Cylindrical
Explanation:
C: Hypermetropia -: When the human eye can see distant object clearly, but cannot see near by distinctly. A convex lens of suitable focal length will correct hypermetropia.
CG PET- 2005
Ray Optics
282480
The focal length of a convex lens is \(10 \mathrm{~cm}\) and its refractive index is 1.5 . If the radius of curvature of one surface in \(7.5 \mathrm{~cm}\), the radius of curvature of the second surface will be
1 \(7.5 \mathrm{~cm}\)
2 \(15.0 \mathrm{~cm}\)
3 \(75 \mathrm{~cm}\)
4 \(5.0 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of convex lens \(=10 \mathrm{~cm}\)
Refractive index \((\mu)=1.5\)
Radius of curvature of \(\mathrm{I}^{\mathrm{st}}\) surface \(\left(\mathrm{R}_1\right)=7.5 \mathrm{~cm}\)
Using power of lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{10}=(1.5-1)\left(\frac{1}{7.5}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{5}=\frac{1}{7.5}-\frac{1}{\mathrm{R}_2} \\
\frac{1}{\mathrm{R}_2}=\frac{1}{7.5}-\frac{1}{5} \\
\frac{1}{\mathrm{R}_2}=\frac{-0.5}{7.5} \\
\mathrm{R}_2=15 \mathrm{~cm}
\end{aligned}\)
CG PET- 2004
Ray Optics
282481
A plano-convex lens of focal length \(30 \mathrm{~cm}\) has its plane surface silvered. An object is placed \(40 \mathrm{~cm}\) from the lens on the convex side. The distance of the image from the lens is
1 \(18 \mathrm{~cm}\)
2 \(24 \mathrm{~cm}\)
3 \(30 \mathrm{~cm}\)
4 \(40 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of plano convex lens (f) \(=30 \mathrm{~cm}\)
Distance of object \((u)=-40 \mathrm{~cm}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{v}}=\frac{1}{30}-\frac{1}{40} \\
\frac{1}{\mathrm{v}}=\frac{4-3}{120} \\
\mathrm{v}=120 \mathrm{~cm}=\mathrm{u}^{\prime}
\end{aligned}\)
Now, this image act as object for the lens.
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{(-120)} \quad\left[\because \mathrm{u}^{\prime}=-120 \mathrm{~cm}\right] \\
\frac{1}{\mathrm{v}^{\prime}}=\frac{1}{30}+\frac{1}{120} \\
\mathrm{v}^{\prime}=24 \mathrm{~cm}
\end{aligned}\)
BITSAT-2009
Ray Optics
282482
A thin convergent glass lens \(\left(\mu_{\mathrm{g}}=1.5\right)\) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index \(\mu\), it acts as a divergent lens of focal length \(100 \mathrm{~cm}\). The value of \(\mu\) must be
1 \(4 / 3\)
2 \(5 / 3\)
3 \(5 / 4(\mathrm{~d})\)
\(6 / 5\)
Explanation:
B: Given,
Refractive index of air \(\left(\mu_{\mathrm{a}}\right)=1\)
Refractive index of thin convergent lens \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Refractive index of liquid \(=\mu\)
Focal length of divergent lens, \(f=-10 \mathrm{~cm}\)
Power of lens,
\(\mathrm{P}_{\mathrm{c}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
When lens immersed in liquid of refractive index ' \(\mu\) '. Focal length formula -
\(\mathrm{P}_{\mathrm{d}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) / (ii)
\(\frac{P_c}{P_d}=\frac{\left(\frac{\mu_g}{\mu}-1\right)}{\left(\frac{\mu_g}{\mu}-1\right)}\)
\(\frac{5}{-1}=\left(\frac{1.5-1}{\frac{1.5}{\mu}-1}\right) \quad\left\{\therefore \mathrm{P}_{\mathrm{d}}=\frac{100}{\mathrm{f}}=\frac{100}{-100}=-1\right\}\)
\(-5=\frac{0.5 \mu}{1.5-\mu}\)
\(-7.5+5 \mu=0.5 \mu\)
\(\mu=\frac{7.5}{4.5}\)
\(\mu=\frac{5}{3}\)
282479
A person suffering from hypermetropia requires which type of spectacle lenses?
1 Concave
2 Plano- concave
3 Convex
4 Cylindrical
Explanation:
C: Hypermetropia -: When the human eye can see distant object clearly, but cannot see near by distinctly. A convex lens of suitable focal length will correct hypermetropia.
CG PET- 2005
Ray Optics
282480
The focal length of a convex lens is \(10 \mathrm{~cm}\) and its refractive index is 1.5 . If the radius of curvature of one surface in \(7.5 \mathrm{~cm}\), the radius of curvature of the second surface will be
1 \(7.5 \mathrm{~cm}\)
2 \(15.0 \mathrm{~cm}\)
3 \(75 \mathrm{~cm}\)
4 \(5.0 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of convex lens \(=10 \mathrm{~cm}\)
Refractive index \((\mu)=1.5\)
Radius of curvature of \(\mathrm{I}^{\mathrm{st}}\) surface \(\left(\mathrm{R}_1\right)=7.5 \mathrm{~cm}\)
Using power of lens formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{10}=(1.5-1)\left(\frac{1}{7.5}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{5}=\frac{1}{7.5}-\frac{1}{\mathrm{R}_2} \\
\frac{1}{\mathrm{R}_2}=\frac{1}{7.5}-\frac{1}{5} \\
\frac{1}{\mathrm{R}_2}=\frac{-0.5}{7.5} \\
\mathrm{R}_2=15 \mathrm{~cm}
\end{aligned}\)
CG PET- 2004
Ray Optics
282481
A plano-convex lens of focal length \(30 \mathrm{~cm}\) has its plane surface silvered. An object is placed \(40 \mathrm{~cm}\) from the lens on the convex side. The distance of the image from the lens is
1 \(18 \mathrm{~cm}\)
2 \(24 \mathrm{~cm}\)
3 \(30 \mathrm{~cm}\)
4 \(40 \mathrm{~cm}\)
Explanation:
B: Given,
Focal length of plano convex lens (f) \(=30 \mathrm{~cm}\)
Distance of object \((u)=-40 \mathrm{~cm}\)
Using lens formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\
\frac{1}{\mathrm{v}}=\frac{1}{30}-\frac{1}{40} \\
\frac{1}{\mathrm{v}}=\frac{4-3}{120} \\
\mathrm{v}=120 \mathrm{~cm}=\mathrm{u}^{\prime}
\end{aligned}\)
Now, this image act as object for the lens.
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{(-120)} \quad\left[\because \mathrm{u}^{\prime}=-120 \mathrm{~cm}\right] \\
\frac{1}{\mathrm{v}^{\prime}}=\frac{1}{30}+\frac{1}{120} \\
\mathrm{v}^{\prime}=24 \mathrm{~cm}
\end{aligned}\)
BITSAT-2009
Ray Optics
282482
A thin convergent glass lens \(\left(\mu_{\mathrm{g}}=1.5\right)\) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index \(\mu\), it acts as a divergent lens of focal length \(100 \mathrm{~cm}\). The value of \(\mu\) must be
1 \(4 / 3\)
2 \(5 / 3\)
3 \(5 / 4(\mathrm{~d})\)
\(6 / 5\)
Explanation:
B: Given,
Refractive index of air \(\left(\mu_{\mathrm{a}}\right)=1\)
Refractive index of thin convergent lens \(\left(\mu_{\mathrm{g}}\right)=1.5\)
Refractive index of liquid \(=\mu\)
Focal length of divergent lens, \(f=-10 \mathrm{~cm}\)
Power of lens,
\(\mathrm{P}_{\mathrm{c}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
When lens immersed in liquid of refractive index ' \(\mu\) '. Focal length formula -
\(\mathrm{P}_{\mathrm{d}}=\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) / (ii)
\(\frac{P_c}{P_d}=\frac{\left(\frac{\mu_g}{\mu}-1\right)}{\left(\frac{\mu_g}{\mu}-1\right)}\)
\(\frac{5}{-1}=\left(\frac{1.5-1}{\frac{1.5}{\mu}-1}\right) \quad\left\{\therefore \mathrm{P}_{\mathrm{d}}=\frac{100}{\mathrm{f}}=\frac{100}{-100}=-1\right\}\)
\(-5=\frac{0.5 \mu}{1.5-\mu}\)
\(-7.5+5 \mu=0.5 \mu\)
\(\mu=\frac{7.5}{4.5}\)
\(\mu=\frac{5}{3}\)
