282490
A diverging meniscus lens of 1.5 refractive index has concave surface of radii 3 and \(4 \mathrm{~cm}\). The position of the image, if an object is placed \(12 \mathrm{~cm}\) in front of the lens, is
1 \(7 \mathrm{~cm}\)
2 \(-8 \mathrm{~cm}\)
3 \(9 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
B: Given,
Refractive index of meniscus lens \(\mu=1.5\)
Radius of curvature of surface \(1\left(R_1\right)=-3 \mathrm{~cm}\)
Radius of curvature of surface \(2\left(R_2\right)=-4 \mathrm{~cm}\) \(\mathrm{u}=-12 \mathrm{~cm} \quad\{-\mathrm{ve}\) sign because diverging lens \(\}\) Using formula -
\(\begin{array}{r}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\end{array}\)
\(\begin{aligned}
\frac{1}{\mathrm{~V}}-\frac{1}{(-12)}=(1.5-1)\left(\frac{1}{-3}-\frac{1}{(-4)}\right) \\
\frac{1}{\mathrm{~V}}=-\frac{1}{12}-\frac{1}{24} \\
\mathrm{v}=-8 \mathrm{~cm}
\end{aligned}\)
UPSEE - 2014
Ray Optics
282491
A wire mesh consisting of very small squares is viewed at a distance of \(8 \mathrm{~cm}\) through a magnifying converging lens of focal length 10 \(\mathrm{cm}\), kept close to the eye. The magnification produced by the lens is
282492
A lens made of glass whose index of refraction is 1.60 has a focal length of \(+20 \mathrm{~cm}\) in air. Its focal length in water, whose refractive index is 1.33 , will be
1 three times longer than in air
2 two times longer than in air
3 same as in air
4 None of the above
Explanation:
A: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.60\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=1.33\)
Focal length ( \(f\) ) \(=20 \mathrm{~cm}\)
Using lens maker formula-
\(\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Again, \(\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mathrm{u}_{\mathrm{g}}}{\mathrm{u}_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) \& (ii)
\(\begin{aligned}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)} \\
\mathrm{f}^{\prime}=20 \frac{(1.6-1)}{\left(\frac{1.6}{1.33}-1\right)} \\
\mathrm{f}^{\prime}=20 \times \frac{0.6 \times 1.33}{0.27} \\
\mathrm{f}^{\prime} \approx 60 \mathrm{~cm} \\
\mathrm{f}^{\prime} \approx 3 \mathrm{f} \quad\{\because \mathrm{f}=20 \mathrm{~cm}\} \\
&
\end{aligned}\)
Hence, its focal length is three time longer in water.
UPSEE - 2008
Ray Optics
282493
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is \(4 \mathrm{D}\), the power of a cut lens will be
1 \(2 \mathrm{D}\)
2 \(3 \mathrm{D}\)
3 \(4 \mathrm{D}\)
4 \(5 \mathrm{D}\)
Explanation:
A: Using lens maker formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \quad\left\{\begin{array}{c}
\text { here, } \mathrm{R}_1=\mathrm{R} \\
\mathrm{R}_2=-\mathrm{R}
\end{array}\right\} \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right)=\frac{2}{\mathrm{R}}(\mu-1)
\end{aligned}\)
Power of lens, \(P=\frac{1}{f}\)
\(\begin{aligned}
4 D & =\frac{2}{R}(\mu-1) \\
2 D & =\frac{(\mu-1)}{R}
\end{aligned}\)
Now, lens cut along perpendicular axis-
\(\mathrm{R}_1=\mathrm{R} \& \mathrm{R}_2=\infty\)
Again,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{\mu-1}{\mathrm{R}}\right)
\end{aligned}\)
Power of lens, \(P^{\prime}=\frac{1}{f^{\prime}}=\frac{\mu-1}{R} \quad\) [From equation (i)]
\(\mathrm{P}^{\prime}=2 \mathrm{D}\)
282490
A diverging meniscus lens of 1.5 refractive index has concave surface of radii 3 and \(4 \mathrm{~cm}\). The position of the image, if an object is placed \(12 \mathrm{~cm}\) in front of the lens, is
1 \(7 \mathrm{~cm}\)
2 \(-8 \mathrm{~cm}\)
3 \(9 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
B: Given,
Refractive index of meniscus lens \(\mu=1.5\)
Radius of curvature of surface \(1\left(R_1\right)=-3 \mathrm{~cm}\)
Radius of curvature of surface \(2\left(R_2\right)=-4 \mathrm{~cm}\) \(\mathrm{u}=-12 \mathrm{~cm} \quad\{-\mathrm{ve}\) sign because diverging lens \(\}\) Using formula -
\(\begin{array}{r}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\end{array}\)
\(\begin{aligned}
\frac{1}{\mathrm{~V}}-\frac{1}{(-12)}=(1.5-1)\left(\frac{1}{-3}-\frac{1}{(-4)}\right) \\
\frac{1}{\mathrm{~V}}=-\frac{1}{12}-\frac{1}{24} \\
\mathrm{v}=-8 \mathrm{~cm}
\end{aligned}\)
UPSEE - 2014
Ray Optics
282491
A wire mesh consisting of very small squares is viewed at a distance of \(8 \mathrm{~cm}\) through a magnifying converging lens of focal length 10 \(\mathrm{cm}\), kept close to the eye. The magnification produced by the lens is
282492
A lens made of glass whose index of refraction is 1.60 has a focal length of \(+20 \mathrm{~cm}\) in air. Its focal length in water, whose refractive index is 1.33 , will be
1 three times longer than in air
2 two times longer than in air
3 same as in air
4 None of the above
Explanation:
A: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.60\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=1.33\)
Focal length ( \(f\) ) \(=20 \mathrm{~cm}\)
Using lens maker formula-
\(\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Again, \(\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mathrm{u}_{\mathrm{g}}}{\mathrm{u}_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) \& (ii)
\(\begin{aligned}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)} \\
\mathrm{f}^{\prime}=20 \frac{(1.6-1)}{\left(\frac{1.6}{1.33}-1\right)} \\
\mathrm{f}^{\prime}=20 \times \frac{0.6 \times 1.33}{0.27} \\
\mathrm{f}^{\prime} \approx 60 \mathrm{~cm} \\
\mathrm{f}^{\prime} \approx 3 \mathrm{f} \quad\{\because \mathrm{f}=20 \mathrm{~cm}\} \\
&
\end{aligned}\)
Hence, its focal length is three time longer in water.
UPSEE - 2008
Ray Optics
282493
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is \(4 \mathrm{D}\), the power of a cut lens will be
1 \(2 \mathrm{D}\)
2 \(3 \mathrm{D}\)
3 \(4 \mathrm{D}\)
4 \(5 \mathrm{D}\)
Explanation:
A: Using lens maker formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \quad\left\{\begin{array}{c}
\text { here, } \mathrm{R}_1=\mathrm{R} \\
\mathrm{R}_2=-\mathrm{R}
\end{array}\right\} \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right)=\frac{2}{\mathrm{R}}(\mu-1)
\end{aligned}\)
Power of lens, \(P=\frac{1}{f}\)
\(\begin{aligned}
4 D & =\frac{2}{R}(\mu-1) \\
2 D & =\frac{(\mu-1)}{R}
\end{aligned}\)
Now, lens cut along perpendicular axis-
\(\mathrm{R}_1=\mathrm{R} \& \mathrm{R}_2=\infty\)
Again,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{\mu-1}{\mathrm{R}}\right)
\end{aligned}\)
Power of lens, \(P^{\prime}=\frac{1}{f^{\prime}}=\frac{\mu-1}{R} \quad\) [From equation (i)]
\(\mathrm{P}^{\prime}=2 \mathrm{D}\)
282490
A diverging meniscus lens of 1.5 refractive index has concave surface of radii 3 and \(4 \mathrm{~cm}\). The position of the image, if an object is placed \(12 \mathrm{~cm}\) in front of the lens, is
1 \(7 \mathrm{~cm}\)
2 \(-8 \mathrm{~cm}\)
3 \(9 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
B: Given,
Refractive index of meniscus lens \(\mu=1.5\)
Radius of curvature of surface \(1\left(R_1\right)=-3 \mathrm{~cm}\)
Radius of curvature of surface \(2\left(R_2\right)=-4 \mathrm{~cm}\) \(\mathrm{u}=-12 \mathrm{~cm} \quad\{-\mathrm{ve}\) sign because diverging lens \(\}\) Using formula -
\(\begin{array}{r}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\end{array}\)
\(\begin{aligned}
\frac{1}{\mathrm{~V}}-\frac{1}{(-12)}=(1.5-1)\left(\frac{1}{-3}-\frac{1}{(-4)}\right) \\
\frac{1}{\mathrm{~V}}=-\frac{1}{12}-\frac{1}{24} \\
\mathrm{v}=-8 \mathrm{~cm}
\end{aligned}\)
UPSEE - 2014
Ray Optics
282491
A wire mesh consisting of very small squares is viewed at a distance of \(8 \mathrm{~cm}\) through a magnifying converging lens of focal length 10 \(\mathrm{cm}\), kept close to the eye. The magnification produced by the lens is
282492
A lens made of glass whose index of refraction is 1.60 has a focal length of \(+20 \mathrm{~cm}\) in air. Its focal length in water, whose refractive index is 1.33 , will be
1 three times longer than in air
2 two times longer than in air
3 same as in air
4 None of the above
Explanation:
A: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.60\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=1.33\)
Focal length ( \(f\) ) \(=20 \mathrm{~cm}\)
Using lens maker formula-
\(\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Again, \(\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mathrm{u}_{\mathrm{g}}}{\mathrm{u}_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) \& (ii)
\(\begin{aligned}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)} \\
\mathrm{f}^{\prime}=20 \frac{(1.6-1)}{\left(\frac{1.6}{1.33}-1\right)} \\
\mathrm{f}^{\prime}=20 \times \frac{0.6 \times 1.33}{0.27} \\
\mathrm{f}^{\prime} \approx 60 \mathrm{~cm} \\
\mathrm{f}^{\prime} \approx 3 \mathrm{f} \quad\{\because \mathrm{f}=20 \mathrm{~cm}\} \\
&
\end{aligned}\)
Hence, its focal length is three time longer in water.
UPSEE - 2008
Ray Optics
282493
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is \(4 \mathrm{D}\), the power of a cut lens will be
1 \(2 \mathrm{D}\)
2 \(3 \mathrm{D}\)
3 \(4 \mathrm{D}\)
4 \(5 \mathrm{D}\)
Explanation:
A: Using lens maker formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \quad\left\{\begin{array}{c}
\text { here, } \mathrm{R}_1=\mathrm{R} \\
\mathrm{R}_2=-\mathrm{R}
\end{array}\right\} \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right)=\frac{2}{\mathrm{R}}(\mu-1)
\end{aligned}\)
Power of lens, \(P=\frac{1}{f}\)
\(\begin{aligned}
4 D & =\frac{2}{R}(\mu-1) \\
2 D & =\frac{(\mu-1)}{R}
\end{aligned}\)
Now, lens cut along perpendicular axis-
\(\mathrm{R}_1=\mathrm{R} \& \mathrm{R}_2=\infty\)
Again,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{\mu-1}{\mathrm{R}}\right)
\end{aligned}\)
Power of lens, \(P^{\prime}=\frac{1}{f^{\prime}}=\frac{\mu-1}{R} \quad\) [From equation (i)]
\(\mathrm{P}^{\prime}=2 \mathrm{D}\)
282490
A diverging meniscus lens of 1.5 refractive index has concave surface of radii 3 and \(4 \mathrm{~cm}\). The position of the image, if an object is placed \(12 \mathrm{~cm}\) in front of the lens, is
1 \(7 \mathrm{~cm}\)
2 \(-8 \mathrm{~cm}\)
3 \(9 \mathrm{~cm}\)
4 \(10 \mathrm{~cm}\)
Explanation:
B: Given,
Refractive index of meniscus lens \(\mu=1.5\)
Radius of curvature of surface \(1\left(R_1\right)=-3 \mathrm{~cm}\)
Radius of curvature of surface \(2\left(R_2\right)=-4 \mathrm{~cm}\) \(\mathrm{u}=-12 \mathrm{~cm} \quad\{-\mathrm{ve}\) sign because diverging lens \(\}\) Using formula -
\(\begin{array}{r}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\end{array}\)
\(\begin{aligned}
\frac{1}{\mathrm{~V}}-\frac{1}{(-12)}=(1.5-1)\left(\frac{1}{-3}-\frac{1}{(-4)}\right) \\
\frac{1}{\mathrm{~V}}=-\frac{1}{12}-\frac{1}{24} \\
\mathrm{v}=-8 \mathrm{~cm}
\end{aligned}\)
UPSEE - 2014
Ray Optics
282491
A wire mesh consisting of very small squares is viewed at a distance of \(8 \mathrm{~cm}\) through a magnifying converging lens of focal length 10 \(\mathrm{cm}\), kept close to the eye. The magnification produced by the lens is
282492
A lens made of glass whose index of refraction is 1.60 has a focal length of \(+20 \mathrm{~cm}\) in air. Its focal length in water, whose refractive index is 1.33 , will be
1 three times longer than in air
2 two times longer than in air
3 same as in air
4 None of the above
Explanation:
A: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.60\)
Refractive index of water \(\left(\mu_{\mathrm{w}}\right)=1.33\)
Focal length ( \(f\) ) \(=20 \mathrm{~cm}\)
Using lens maker formula-
\(\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Again, \(\frac{1}{\mathrm{f}^{\prime}}=\left(\frac{\mathrm{u}_{\mathrm{g}}}{\mathrm{u}_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Equation (i) \& (ii)
\(\begin{aligned}
\frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\left(\mu_{\mathrm{g}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)} \\
\mathrm{f}^{\prime}=20 \frac{(1.6-1)}{\left(\frac{1.6}{1.33}-1\right)} \\
\mathrm{f}^{\prime}=20 \times \frac{0.6 \times 1.33}{0.27} \\
\mathrm{f}^{\prime} \approx 60 \mathrm{~cm} \\
\mathrm{f}^{\prime} \approx 3 \mathrm{f} \quad\{\because \mathrm{f}=20 \mathrm{~cm}\} \\
&
\end{aligned}\)
Hence, its focal length is three time longer in water.
UPSEE - 2008
Ray Optics
282493
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is \(4 \mathrm{D}\), the power of a cut lens will be
1 \(2 \mathrm{D}\)
2 \(3 \mathrm{D}\)
3 \(4 \mathrm{D}\)
4 \(5 \mathrm{D}\)
Explanation:
A: Using lens maker formula-
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \quad\left\{\begin{array}{c}
\text { here, } \mathrm{R}_1=\mathrm{R} \\
\mathrm{R}_2=-\mathrm{R}
\end{array}\right\} \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right)=\frac{2}{\mathrm{R}}(\mu-1)
\end{aligned}\)
Power of lens, \(P=\frac{1}{f}\)
\(\begin{aligned}
4 D & =\frac{2}{R}(\mu-1) \\
2 D & =\frac{(\mu-1)}{R}
\end{aligned}\)
Now, lens cut along perpendicular axis-
\(\mathrm{R}_1=\mathrm{R} \& \mathrm{R}_2=\infty\)
Again,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right) \\
\frac{1}{\mathrm{f}}=\left(\frac{\mu-1}{\mathrm{R}}\right)
\end{aligned}\)
Power of lens, \(P^{\prime}=\frac{1}{f^{\prime}}=\frac{\mu-1}{R} \quad\) [From equation (i)]
\(\mathrm{P}^{\prime}=2 \mathrm{D}\)
