Answer
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Hint:- The above problem can be solved using the formula derived from the object image and focal distance relationship formulas of the convex lens and the convex mirror of the same focal length. The formula for the focal length of the lens and the mirror is used.
Useful formula:
The distance of the value $d$;
\[\dfrac{1}{d} = \dfrac{1}{f}\]
Where, the $d$ is the distance between the convex lens and convex mirror, $f$ is the focal length of the convex mirror.
Complete step by step solution:
The data given in the problem is;
Focal length of the convex lens is, ${f_1} = 20\,\,cm$.
Focal length of the convex mirror is, ${f_2} = 20\,\,cm$
Distance of the image placed from the image is, ${u_1} = 30\,\,cm$
At convex lens;
$\dfrac{1}{{{f_1}}} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}}$
Substituting the value of focal length and the object distance from the lens
$
\dfrac{1}{{20}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{30}} \\
\dfrac{1}{{20}} - \dfrac{1}{{30}} = \dfrac{1}{{{d_i}}} \\
$
Where, ${d_i}$ denotes the distance of the image at convex lens.
\[\dfrac{1}{{{d_i}}} = \dfrac{1}{{60}}\]
At convex mirror:
$\dfrac{1}{{{f_2}}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{{u_o}}}$
Substitutes the values of the focal length and the image distance;
$
\dfrac{1}{{20}} = \dfrac{1}{{60}} + \dfrac{1}{{{d_o}}} \\
\dfrac{1}{{20}} - \dfrac{1}{{60}} = \dfrac{1}{{{d_o}}} \\
$
Where, ${d_o}$ denotes the distance of the object at convex mirror.
\[\dfrac{1}{{{d_o}}} = \dfrac{1}{{30}}\]
The distance of the value $d$;
\[\dfrac{1}{d} = \dfrac{1}{f}\]
That is
\[
\dfrac{1}{d} = \dfrac{1}{f} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{{d_o}}} \\
\dfrac{1}{d} = \dfrac{1}{{60}} + \dfrac{1}{{30}} \\
\dfrac{1}{d} = \dfrac{{90}}{{1800}} \\
d = 20\,\,cm \\
\]
Therefore, the value of the $d$ is 20 cm.
Hence the option (D), \[d = 20\,\,cm\] is the correct answer.
Note: Image distance denotes that when the image is created then the distance between pole and image is known image distance. Focal length is the interval between pole and the principal focus of the mirror.
Useful formula:
The distance of the value $d$;
\[\dfrac{1}{d} = \dfrac{1}{f}\]
Where, the $d$ is the distance between the convex lens and convex mirror, $f$ is the focal length of the convex mirror.
Complete step by step solution:
The data given in the problem is;
Focal length of the convex lens is, ${f_1} = 20\,\,cm$.
Focal length of the convex mirror is, ${f_2} = 20\,\,cm$
Distance of the image placed from the image is, ${u_1} = 30\,\,cm$
At convex lens;
$\dfrac{1}{{{f_1}}} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}}$
Substituting the value of focal length and the object distance from the lens
$
\dfrac{1}{{20}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{30}} \\
\dfrac{1}{{20}} - \dfrac{1}{{30}} = \dfrac{1}{{{d_i}}} \\
$
Where, ${d_i}$ denotes the distance of the image at convex lens.
\[\dfrac{1}{{{d_i}}} = \dfrac{1}{{60}}\]
At convex mirror:
$\dfrac{1}{{{f_2}}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{{u_o}}}$
Substitutes the values of the focal length and the image distance;
$
\dfrac{1}{{20}} = \dfrac{1}{{60}} + \dfrac{1}{{{d_o}}} \\
\dfrac{1}{{20}} - \dfrac{1}{{60}} = \dfrac{1}{{{d_o}}} \\
$
Where, ${d_o}$ denotes the distance of the object at convex mirror.
\[\dfrac{1}{{{d_o}}} = \dfrac{1}{{30}}\]
The distance of the value $d$;
\[\dfrac{1}{d} = \dfrac{1}{f}\]
That is
\[
\dfrac{1}{d} = \dfrac{1}{f} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{{d_o}}} \\
\dfrac{1}{d} = \dfrac{1}{{60}} + \dfrac{1}{{30}} \\
\dfrac{1}{d} = \dfrac{{90}}{{1800}} \\
d = 20\,\,cm \\
\]
Therefore, the value of the $d$ is 20 cm.
Hence the option (D), \[d = 20\,\,cm\] is the correct answer.
Note: Image distance denotes that when the image is created then the distance between pole and image is known image distance. Focal length is the interval between pole and the principal focus of the mirror.
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