Solveeit Logo

Question

Question: A convex lens and convex mirror are placed co axially and separated by distance d. The focal length ...

A convex lens and convex mirror are placed co axially and separated by distance d. The focal length of both is 20cm20\,\,cm each. A point object is placed at a distance 30cm30\,\,cm from the lens as shown. Then the value of d so that image on the object itself is

A) 10cm10\,\,cm
B) 60cm60\,\,cm
C) 30cm30\,\,cm
D) 20cm20\,\,cm

Explanation

Solution

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 dd;
1d=1f\dfrac{1}{d} = \dfrac{1}{f}
Where, the dd is the distance between the convex lens and convex mirror, ff 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, f1=20cm{f_1} = 20\,\,cm.
Focal length of the convex mirror is, f2=20cm{f_2} = 20\,\,cm
Distance of the image placed from the image is, u1=30cm{u_1} = 30\,\,cm

At convex lens;
1f1=1v1+1u1\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
120=1di+130 120130=1di  \dfrac{1}{{20}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{30}} \\\ \dfrac{1}{{20}} - \dfrac{1}{{30}} = \dfrac{1}{{{d_i}}} \\\
Where, di{d_i} denotes the distance of the image at convex lens.
1di=160\dfrac{1}{{{d_i}}} = \dfrac{1}{{60}}

At convex mirror:
1f2=1di+1uo\dfrac{1}{{{f_2}}} = \dfrac{1}{{{d_i}}} + \dfrac{1}{{{u_o}}}
Substitutes the values of the focal length and the image distance;
120=160+1do 120160=1do  \dfrac{1}{{20}} = \dfrac{1}{{60}} + \dfrac{1}{{{d_o}}} \\\ \dfrac{1}{{20}} - \dfrac{1}{{60}} = \dfrac{1}{{{d_o}}} \\\
Where, do{d_o} denotes the distance of the object at convex mirror.
1do=130\dfrac{1}{{{d_o}}} = \dfrac{1}{{30}}

The distance of the value dd;
1d=1f\dfrac{1}{d} = \dfrac{1}{f}
That is

1d=1f=1di+1do 1d=160+130 1d=901800 d=20cm  \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 dd is 20 cm.

Hence the option (D), d=20cmd = 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.