282384
A plano-convex lens is made from glass of refractive index 1.5 . The radius of curvature of its curved surface is ' \(R\) '. Its focal length is
1 \(\mathrm{R} / 2\)
2 \(2 \mathrm{R}\)
3 \(1.5 \mathrm{R}\)
4 \(\mathrm{R}\)
Explanation:
B: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
\(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\infty\)
We know that,
From lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{aligned}\)
\begin{tabular}{l}
\(\frac{1}{\mathrm{f}}=\frac{0.5}{\mathrm{R}}\) \\
\(\frac{1}{\mathrm{f}}=\frac{1}{2 \mathrm{R}}\) \\
\(\therefore \quad \mathrm{f}=2 \mathrm{R}\)
\end{tabular}
MHT-CET 2020
Ray Optics
282385
The size of the real image produced by a convex lens of focal length \(F\) is \(m\) times the size of the object. The image distance from the lens is
1 \(\mathrm{F}(\mathrm{m}-1)\)
2 \(\mathrm{F}(\mathrm{m}+1)\)
3 \(\frac{(\mathrm{m}-1)}{\mathrm{F}}\)
4 \(\frac{\mathrm{F}}{(\mathrm{m}-1)}\)
Explanation:
B: Given, focal length \(=\mathrm{F}\)
We know that,
Magnification of convex lens, \(m=\frac{\mathrm{V}}{\mathrm{u}}\)
\(\mathrm{v}=-\mathrm{mu} \quad\{-\mathrm{ve} \text { is for real image }\}\)
From lens formula,
\(\begin{aligned}
\frac{1}{F}=\frac{1}{V}-\frac{1}{u} \\
\frac{1}{F}=\frac{1}{V}+\frac{1}{\mathrm{v} / \mathrm{m}} \\
\frac{1}{\mathrm{~F}}=\frac{(1+\mathrm{m})}{\mathrm{V}} \\
\mathrm{V}=\mathrm{F}(1+\mathrm{m})
\end{aligned}\)
MHT-CET 2020
Ray Optics
282386
The figure shows equiconvex lens of focal length ' \(f\) '. If the lens is cut along \(A B\), the focal length of each half will be
282384
A plano-convex lens is made from glass of refractive index 1.5 . The radius of curvature of its curved surface is ' \(R\) '. Its focal length is
1 \(\mathrm{R} / 2\)
2 \(2 \mathrm{R}\)
3 \(1.5 \mathrm{R}\)
4 \(\mathrm{R}\)
Explanation:
B: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
\(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\infty\)
We know that,
From lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{aligned}\)
\begin{tabular}{l}
\(\frac{1}{\mathrm{f}}=\frac{0.5}{\mathrm{R}}\) \\
\(\frac{1}{\mathrm{f}}=\frac{1}{2 \mathrm{R}}\) \\
\(\therefore \quad \mathrm{f}=2 \mathrm{R}\)
\end{tabular}
MHT-CET 2020
Ray Optics
282385
The size of the real image produced by a convex lens of focal length \(F\) is \(m\) times the size of the object. The image distance from the lens is
1 \(\mathrm{F}(\mathrm{m}-1)\)
2 \(\mathrm{F}(\mathrm{m}+1)\)
3 \(\frac{(\mathrm{m}-1)}{\mathrm{F}}\)
4 \(\frac{\mathrm{F}}{(\mathrm{m}-1)}\)
Explanation:
B: Given, focal length \(=\mathrm{F}\)
We know that,
Magnification of convex lens, \(m=\frac{\mathrm{V}}{\mathrm{u}}\)
\(\mathrm{v}=-\mathrm{mu} \quad\{-\mathrm{ve} \text { is for real image }\}\)
From lens formula,
\(\begin{aligned}
\frac{1}{F}=\frac{1}{V}-\frac{1}{u} \\
\frac{1}{F}=\frac{1}{V}+\frac{1}{\mathrm{v} / \mathrm{m}} \\
\frac{1}{\mathrm{~F}}=\frac{(1+\mathrm{m})}{\mathrm{V}} \\
\mathrm{V}=\mathrm{F}(1+\mathrm{m})
\end{aligned}\)
MHT-CET 2020
Ray Optics
282386
The figure shows equiconvex lens of focal length ' \(f\) '. If the lens is cut along \(A B\), the focal length of each half will be
282384
A plano-convex lens is made from glass of refractive index 1.5 . The radius of curvature of its curved surface is ' \(R\) '. Its focal length is
1 \(\mathrm{R} / 2\)
2 \(2 \mathrm{R}\)
3 \(1.5 \mathrm{R}\)
4 \(\mathrm{R}\)
Explanation:
B: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
\(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\infty\)
We know that,
From lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{aligned}\)
\begin{tabular}{l}
\(\frac{1}{\mathrm{f}}=\frac{0.5}{\mathrm{R}}\) \\
\(\frac{1}{\mathrm{f}}=\frac{1}{2 \mathrm{R}}\) \\
\(\therefore \quad \mathrm{f}=2 \mathrm{R}\)
\end{tabular}
MHT-CET 2020
Ray Optics
282385
The size of the real image produced by a convex lens of focal length \(F\) is \(m\) times the size of the object. The image distance from the lens is
1 \(\mathrm{F}(\mathrm{m}-1)\)
2 \(\mathrm{F}(\mathrm{m}+1)\)
3 \(\frac{(\mathrm{m}-1)}{\mathrm{F}}\)
4 \(\frac{\mathrm{F}}{(\mathrm{m}-1)}\)
Explanation:
B: Given, focal length \(=\mathrm{F}\)
We know that,
Magnification of convex lens, \(m=\frac{\mathrm{V}}{\mathrm{u}}\)
\(\mathrm{v}=-\mathrm{mu} \quad\{-\mathrm{ve} \text { is for real image }\}\)
From lens formula,
\(\begin{aligned}
\frac{1}{F}=\frac{1}{V}-\frac{1}{u} \\
\frac{1}{F}=\frac{1}{V}+\frac{1}{\mathrm{v} / \mathrm{m}} \\
\frac{1}{\mathrm{~F}}=\frac{(1+\mathrm{m})}{\mathrm{V}} \\
\mathrm{V}=\mathrm{F}(1+\mathrm{m})
\end{aligned}\)
MHT-CET 2020
Ray Optics
282386
The figure shows equiconvex lens of focal length ' \(f\) '. If the lens is cut along \(A B\), the focal length of each half will be
282384
A plano-convex lens is made from glass of refractive index 1.5 . The radius of curvature of its curved surface is ' \(R\) '. Its focal length is
1 \(\mathrm{R} / 2\)
2 \(2 \mathrm{R}\)
3 \(1.5 \mathrm{R}\)
4 \(\mathrm{R}\)
Explanation:
B: Given,
Refractive index of glass \(\left(\mu_{\mathrm{g}}\right)=1.5\),
\(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\infty\)
We know that,
From lens makers formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)
\end{aligned}\)
\begin{tabular}{l}
\(\frac{1}{\mathrm{f}}=\frac{0.5}{\mathrm{R}}\) \\
\(\frac{1}{\mathrm{f}}=\frac{1}{2 \mathrm{R}}\) \\
\(\therefore \quad \mathrm{f}=2 \mathrm{R}\)
\end{tabular}
MHT-CET 2020
Ray Optics
282385
The size of the real image produced by a convex lens of focal length \(F\) is \(m\) times the size of the object. The image distance from the lens is
1 \(\mathrm{F}(\mathrm{m}-1)\)
2 \(\mathrm{F}(\mathrm{m}+1)\)
3 \(\frac{(\mathrm{m}-1)}{\mathrm{F}}\)
4 \(\frac{\mathrm{F}}{(\mathrm{m}-1)}\)
Explanation:
B: Given, focal length \(=\mathrm{F}\)
We know that,
Magnification of convex lens, \(m=\frac{\mathrm{V}}{\mathrm{u}}\)
\(\mathrm{v}=-\mathrm{mu} \quad\{-\mathrm{ve} \text { is for real image }\}\)
From lens formula,
\(\begin{aligned}
\frac{1}{F}=\frac{1}{V}-\frac{1}{u} \\
\frac{1}{F}=\frac{1}{V}+\frac{1}{\mathrm{v} / \mathrm{m}} \\
\frac{1}{\mathrm{~F}}=\frac{(1+\mathrm{m})}{\mathrm{V}} \\
\mathrm{V}=\mathrm{F}(1+\mathrm{m})
\end{aligned}\)
MHT-CET 2020
Ray Optics
282386
The figure shows equiconvex lens of focal length ' \(f\) '. If the lens is cut along \(A B\), the focal length of each half will be
