Question

A convex lens of focal length 20cm is placed coaxially with a convex mirror of radius of curvature 20cm. The two are kept at 15cm from each other. A point object placed 40cm in front of the convex lens. Find the position of the image formed by this combination.

Answer 1

Hint: The image formed by the convex lens on the right side of it will act as a virtual object for the convex mirror. The focal length of the convex mirror will be the half of its centre of curvature. The distance of the virtual object from the mirror will be the distance between two magnitudes. Refer to the solution below.

Formula used: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

Complete step-by-step answer:


The distance of the object from the lens as per given in the question is $u = - 40cm$. (We will take the negative sign as the object is put on the left side of the lens.)
Focal length of the lens is given in the question i.e. $f = 20cm$.
In order to find the distance of the image formed by the lens, we will use the Lens formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Thus, we can say that-
$
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{ - 40}} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} - \dfrac{1}{{40}} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{2 - 1}}{{40}} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{{40}} \\
    \\
   \Rightarrow v = 40cm \\
$
Since the value of v is positive, we can say that the image will be formed at the right side of the lens at ${I_2}$ which is behind the convex mirror.
Since there is a convex mirror after the convex lens, we have to find the final image formed by the combination of convex lens and mirror.
Now, the image formed at ${I_2}$ will act as a virtual object since it is at the back side of the mirror.
The convex mirror will form an image ${I_1}$.
In order to find the distance of the image formed by the mirror, we will use the Mirror formula $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$. Thus, we can say that-
Focal length of the mirror f’ will be the half of its radius of curvature (C). Hence,
$
   \Rightarrow f = \dfrac{C}{2} \\
    \\
   \Rightarrow f' = \dfrac{{20}}{2} \\
    \\
   \Rightarrow f' = 10cm \\
$
The distance of the virtual object from the mirror u’ (image formed by the lens) will be the difference between the distance of the image ${I_2}$ from the lens and the distance between the lens and the mirror-
$
   \Rightarrow u' = 40cm - 15cm \\
    \\
   \Rightarrow u' = 25cm \\
$
The distance of the final image formed by the mirror (v’)-
Applying the Mirror formula-
$
   \Rightarrow \dfrac{1}{{f'}} = \dfrac{1}{{v'}} + \dfrac{1}{{u'}} \\
    \\
   \Rightarrow \dfrac{1}{{10}} = \dfrac{1}{{v'}} + \dfrac{1}{{25}} \\
    \\
   \Rightarrow \dfrac{1}{{v'}} = \dfrac{1}{{10}} - \dfrac{1}{{25}} \\
    \\
   \Rightarrow \dfrac{1}{{v'}} = \dfrac{{25 - 10}}{{250}} \\
    \\
   \Rightarrow \dfrac{1}{{v'}} = \dfrac{{15}}{{250}} \\
    \\
   \Rightarrow v' = \dfrac{{250}}{{15}} \\
    \\
   \Rightarrow v' = 16.67cm \\
$
Hence, the distance of the final image formed by the mirror is 16.67cm from the mirror.


Note: Two spherical surfaces usually consist of an optical lens. The lens is called a biconvex lens or merely a convex lens because both surfaces are bent internally. These lens forms will concentrate a light beam from outside and focus it on the other side. Convex Mirror is a curved mirror in the direction of the reflexive surface. The bulging surface reflects exterior light which is not intended for light focusing. Both mirrors form a composite picture as the fixation point (F) and the middle of the bend (2F), which cannot be seen in the mirror.