282380
A biconvex lens \(\left(R_1=R_2=20 \mathrm{~cm}\right)\) has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens \(=1.5\) )
1 \(-20 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(-40 \mathrm{~cm}\)
Explanation:
D: Given, \(\mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}\)
Focal length of lens \(=\) focal length of concave mirror.
\(\mathrm{R}\) of concave mirror \(=\) ?
\(\mu_{\mathrm{g}}=1.5\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right) \\
\frac{1}{\mathrm{f}}=\frac{0.5 \times 2}{20} \\
\mathrm{f}=20 \mathrm{~cm}=\text { focal length of mirror }
\end{aligned}\)
Hence, Radius of curvature of concave mirror
\(=2 \times \mathrm{f}=2 \times(-20)=-40 \mathrm{~cm}\)
(Negative sign for concave mirror)
МНT-CET 2020
Ray Optics
282363
The focal length of a biconvex lens, made of glass, of equal radii is \(f\). If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as \(3 / 2\) and \(4 / 3\), respectively)
1 \(2 \mathrm{f}\)
2 \(4 \mathrm{f}\)
3 \((5 / 3) \mathrm{f}\)
4 \((7 / 4) \mathrm{f}\)
Explanation:
B: Radius are equal,
\(\begin{aligned}
R_1=R_2=r \\
{ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}
\end{aligned}\)
According to lens makers formula,
\(\therefore \quad \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Here, \(\mathrm{R}_1=\mathrm{R}\) and \(\mathrm{R}_2=-\mathrm{R}\)
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{f}} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=\frac{1}{2} \times \frac{2}{\mathrm{R}}=\frac{1}{\mathrm{R}} \\
=\mathrm{R}
\end{aligned}\)
When the lens is dipped in the water,
\(\begin{aligned}
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{3 / 2-1}{4 / 3-1}\right) \times \frac{2}{\mathrm{R}}=\left(\frac{9}{8}-1\right) \times \frac{2}{\mathrm{R}} \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \times \frac{2}{\mathrm{R}}=\frac{1}{4 \mathrm{R}} \\
\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}
\end{aligned}\)
TS EAMCET 06.08.2021
Ray Optics
282364
Two lenses of power \(-1.6 \quad D\) and +2.1D respectively are placed in contact. The focal length of the combinations is
282366
The Radii of curvature of the faces of a double convex lens are \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\). Its focal length is \(12 \mathrm{~cm}\). What is the refractive index of material of lens?
1 1.33
2 1.62
3 1.50
4 2.42
Explanation:
C: Given,
Radius of curvature of first face \(\left(R_1\right)=10 \mathrm{~cm}\)
Radius of curvature of second face \(\left(R_2\right)=-15 \mathrm{~cm}\)
Focal length \((\mathrm{f})=12 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{3+2}{30}\right) \\
\mu-1=\frac{6}{12}=\frac{1}{2} \\
\mu=\frac{1}{2}+1=3 / 2 \\
\mu=1.5
\end{aligned}\)
or
\(\mu=1.5\)
282380
A biconvex lens \(\left(R_1=R_2=20 \mathrm{~cm}\right)\) has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens \(=1.5\) )
1 \(-20 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(-40 \mathrm{~cm}\)
Explanation:
D: Given, \(\mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}\)
Focal length of lens \(=\) focal length of concave mirror.
\(\mathrm{R}\) of concave mirror \(=\) ?
\(\mu_{\mathrm{g}}=1.5\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right) \\
\frac{1}{\mathrm{f}}=\frac{0.5 \times 2}{20} \\
\mathrm{f}=20 \mathrm{~cm}=\text { focal length of mirror }
\end{aligned}\)
Hence, Radius of curvature of concave mirror
\(=2 \times \mathrm{f}=2 \times(-20)=-40 \mathrm{~cm}\)
(Negative sign for concave mirror)
МНT-CET 2020
Ray Optics
282363
The focal length of a biconvex lens, made of glass, of equal radii is \(f\). If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as \(3 / 2\) and \(4 / 3\), respectively)
1 \(2 \mathrm{f}\)
2 \(4 \mathrm{f}\)
3 \((5 / 3) \mathrm{f}\)
4 \((7 / 4) \mathrm{f}\)
Explanation:
B: Radius are equal,
\(\begin{aligned}
R_1=R_2=r \\
{ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}
\end{aligned}\)
According to lens makers formula,
\(\therefore \quad \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Here, \(\mathrm{R}_1=\mathrm{R}\) and \(\mathrm{R}_2=-\mathrm{R}\)
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{f}} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=\frac{1}{2} \times \frac{2}{\mathrm{R}}=\frac{1}{\mathrm{R}} \\
=\mathrm{R}
\end{aligned}\)
When the lens is dipped in the water,
\(\begin{aligned}
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{3 / 2-1}{4 / 3-1}\right) \times \frac{2}{\mathrm{R}}=\left(\frac{9}{8}-1\right) \times \frac{2}{\mathrm{R}} \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \times \frac{2}{\mathrm{R}}=\frac{1}{4 \mathrm{R}} \\
\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}
\end{aligned}\)
TS EAMCET 06.08.2021
Ray Optics
282364
Two lenses of power \(-1.6 \quad D\) and +2.1D respectively are placed in contact. The focal length of the combinations is
282366
The Radii of curvature of the faces of a double convex lens are \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\). Its focal length is \(12 \mathrm{~cm}\). What is the refractive index of material of lens?
1 1.33
2 1.62
3 1.50
4 2.42
Explanation:
C: Given,
Radius of curvature of first face \(\left(R_1\right)=10 \mathrm{~cm}\)
Radius of curvature of second face \(\left(R_2\right)=-15 \mathrm{~cm}\)
Focal length \((\mathrm{f})=12 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{3+2}{30}\right) \\
\mu-1=\frac{6}{12}=\frac{1}{2} \\
\mu=\frac{1}{2}+1=3 / 2 \\
\mu=1.5
\end{aligned}\)
or
\(\mu=1.5\)
282380
A biconvex lens \(\left(R_1=R_2=20 \mathrm{~cm}\right)\) has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens \(=1.5\) )
1 \(-20 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(-40 \mathrm{~cm}\)
Explanation:
D: Given, \(\mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}\)
Focal length of lens \(=\) focal length of concave mirror.
\(\mathrm{R}\) of concave mirror \(=\) ?
\(\mu_{\mathrm{g}}=1.5\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right) \\
\frac{1}{\mathrm{f}}=\frac{0.5 \times 2}{20} \\
\mathrm{f}=20 \mathrm{~cm}=\text { focal length of mirror }
\end{aligned}\)
Hence, Radius of curvature of concave mirror
\(=2 \times \mathrm{f}=2 \times(-20)=-40 \mathrm{~cm}\)
(Negative sign for concave mirror)
МНT-CET 2020
Ray Optics
282363
The focal length of a biconvex lens, made of glass, of equal radii is \(f\). If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as \(3 / 2\) and \(4 / 3\), respectively)
1 \(2 \mathrm{f}\)
2 \(4 \mathrm{f}\)
3 \((5 / 3) \mathrm{f}\)
4 \((7 / 4) \mathrm{f}\)
Explanation:
B: Radius are equal,
\(\begin{aligned}
R_1=R_2=r \\
{ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}
\end{aligned}\)
According to lens makers formula,
\(\therefore \quad \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Here, \(\mathrm{R}_1=\mathrm{R}\) and \(\mathrm{R}_2=-\mathrm{R}\)
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{f}} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=\frac{1}{2} \times \frac{2}{\mathrm{R}}=\frac{1}{\mathrm{R}} \\
=\mathrm{R}
\end{aligned}\)
When the lens is dipped in the water,
\(\begin{aligned}
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{3 / 2-1}{4 / 3-1}\right) \times \frac{2}{\mathrm{R}}=\left(\frac{9}{8}-1\right) \times \frac{2}{\mathrm{R}} \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \times \frac{2}{\mathrm{R}}=\frac{1}{4 \mathrm{R}} \\
\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}
\end{aligned}\)
TS EAMCET 06.08.2021
Ray Optics
282364
Two lenses of power \(-1.6 \quad D\) and +2.1D respectively are placed in contact. The focal length of the combinations is
282366
The Radii of curvature of the faces of a double convex lens are \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\). Its focal length is \(12 \mathrm{~cm}\). What is the refractive index of material of lens?
1 1.33
2 1.62
3 1.50
4 2.42
Explanation:
C: Given,
Radius of curvature of first face \(\left(R_1\right)=10 \mathrm{~cm}\)
Radius of curvature of second face \(\left(R_2\right)=-15 \mathrm{~cm}\)
Focal length \((\mathrm{f})=12 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{3+2}{30}\right) \\
\mu-1=\frac{6}{12}=\frac{1}{2} \\
\mu=\frac{1}{2}+1=3 / 2 \\
\mu=1.5
\end{aligned}\)
or
\(\mu=1.5\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ray Optics
282380
A biconvex lens \(\left(R_1=R_2=20 \mathrm{~cm}\right)\) has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens \(=1.5\) )
1 \(-20 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(-40 \mathrm{~cm}\)
Explanation:
D: Given, \(\mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}\)
Focal length of lens \(=\) focal length of concave mirror.
\(\mathrm{R}\) of concave mirror \(=\) ?
\(\mu_{\mathrm{g}}=1.5\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right) \\
\frac{1}{\mathrm{f}}=\frac{0.5 \times 2}{20} \\
\mathrm{f}=20 \mathrm{~cm}=\text { focal length of mirror }
\end{aligned}\)
Hence, Radius of curvature of concave mirror
\(=2 \times \mathrm{f}=2 \times(-20)=-40 \mathrm{~cm}\)
(Negative sign for concave mirror)
МНT-CET 2020
Ray Optics
282363
The focal length of a biconvex lens, made of glass, of equal radii is \(f\). If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as \(3 / 2\) and \(4 / 3\), respectively)
1 \(2 \mathrm{f}\)
2 \(4 \mathrm{f}\)
3 \((5 / 3) \mathrm{f}\)
4 \((7 / 4) \mathrm{f}\)
Explanation:
B: Radius are equal,
\(\begin{aligned}
R_1=R_2=r \\
{ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}
\end{aligned}\)
According to lens makers formula,
\(\therefore \quad \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Here, \(\mathrm{R}_1=\mathrm{R}\) and \(\mathrm{R}_2=-\mathrm{R}\)
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{f}} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=\frac{1}{2} \times \frac{2}{\mathrm{R}}=\frac{1}{\mathrm{R}} \\
=\mathrm{R}
\end{aligned}\)
When the lens is dipped in the water,
\(\begin{aligned}
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{3 / 2-1}{4 / 3-1}\right) \times \frac{2}{\mathrm{R}}=\left(\frac{9}{8}-1\right) \times \frac{2}{\mathrm{R}} \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \times \frac{2}{\mathrm{R}}=\frac{1}{4 \mathrm{R}} \\
\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}
\end{aligned}\)
TS EAMCET 06.08.2021
Ray Optics
282364
Two lenses of power \(-1.6 \quad D\) and +2.1D respectively are placed in contact. The focal length of the combinations is
282366
The Radii of curvature of the faces of a double convex lens are \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\). Its focal length is \(12 \mathrm{~cm}\). What is the refractive index of material of lens?
1 1.33
2 1.62
3 1.50
4 2.42
Explanation:
C: Given,
Radius of curvature of first face \(\left(R_1\right)=10 \mathrm{~cm}\)
Radius of curvature of second face \(\left(R_2\right)=-15 \mathrm{~cm}\)
Focal length \((\mathrm{f})=12 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{3+2}{30}\right) \\
\mu-1=\frac{6}{12}=\frac{1}{2} \\
\mu=\frac{1}{2}+1=3 / 2 \\
\mu=1.5
\end{aligned}\)
or
\(\mu=1.5\)
282380
A biconvex lens \(\left(R_1=R_2=20 \mathrm{~cm}\right)\) has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens \(=1.5\) )
1 \(-20 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\)
3 \(20 \mathrm{~cm}\)
4 \(-40 \mathrm{~cm}\)
Explanation:
D: Given, \(\mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}\)
Focal length of lens \(=\) focal length of concave mirror.
\(\mathrm{R}\) of concave mirror \(=\) ?
\(\mu_{\mathrm{g}}=1.5\)
We know that,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right) \\
\frac{1}{\mathrm{f}}=\frac{0.5 \times 2}{20} \\
\mathrm{f}=20 \mathrm{~cm}=\text { focal length of mirror }
\end{aligned}\)
Hence, Radius of curvature of concave mirror
\(=2 \times \mathrm{f}=2 \times(-20)=-40 \mathrm{~cm}\)
(Negative sign for concave mirror)
МНT-CET 2020
Ray Optics
282363
The focal length of a biconvex lens, made of glass, of equal radii is \(f\). If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as \(3 / 2\) and \(4 / 3\), respectively)
1 \(2 \mathrm{f}\)
2 \(4 \mathrm{f}\)
3 \((5 / 3) \mathrm{f}\)
4 \((7 / 4) \mathrm{f}\)
Explanation:
B: Radius are equal,
\(\begin{aligned}
R_1=R_2=r \\
{ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}
\end{aligned}\)
According to lens makers formula,
\(\therefore \quad \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Here, \(\mathrm{R}_1=\mathrm{R}\) and \(\mathrm{R}_2=-\mathrm{R}\)
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{f}} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=\frac{1}{2} \times \frac{2}{\mathrm{R}}=\frac{1}{\mathrm{R}} \\
=\mathrm{R}
\end{aligned}\)
When the lens is dipped in the water,
\(\begin{aligned}
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{3 / 2-1}{4 / 3-1}\right) \times \frac{2}{\mathrm{R}}=\left(\frac{9}{8}-1\right) \times \frac{2}{\mathrm{R}} \\
\frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \times \frac{2}{\mathrm{R}}=\frac{1}{4 \mathrm{R}} \\
\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}
\end{aligned}\)
TS EAMCET 06.08.2021
Ray Optics
282364
Two lenses of power \(-1.6 \quad D\) and +2.1D respectively are placed in contact. The focal length of the combinations is
282366
The Radii of curvature of the faces of a double convex lens are \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\). Its focal length is \(12 \mathrm{~cm}\). What is the refractive index of material of lens?
1 1.33
2 1.62
3 1.50
4 2.42
Explanation:
C: Given,
Radius of curvature of first face \(\left(R_1\right)=10 \mathrm{~cm}\)
Radius of curvature of second face \(\left(R_2\right)=-15 \mathrm{~cm}\)
Focal length \((\mathrm{f})=12 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right) \\
\frac{1}{12}=(\mu-1)\left(\frac{3+2}{30}\right) \\
\mu-1=\frac{6}{12}=\frac{1}{2} \\
\mu=\frac{1}{2}+1=3 / 2 \\
\mu=1.5
\end{aligned}\)
or
\(\mu=1.5\)