Question

A simple camera with a converging lens of focal length of 60 mm is focused on distant objects. Now, to focus the camera on a nearby object placed at a distance of 1.5 m, find the change in the distance between the film and the lens.
A) The distance must be decreased by 2.5 mm.
B) The distance must be increased by 2.5 mm.
C) The distance remains the same but the aperture must be increased by a factor of 2.5 mm.
D) The distance remains the same but the aperture must be decreased by a factor of 2.5 mm.

Answer 1

The distance between the film and the lens refers to the position of the image from the converging lens. Here, the camera is said to have a converging lens i.e., it has a convex lens. For a convex lens, the image of a distant object will be formed on its focal point.

Formula Used: The thin lens formula is given by, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ where, $v$ is the position of the image from the lens, $u$ is the position of the object from the lens and $f$ is the focal length of the lens.

Complete step by step answer:
Step 1: List the parameters known from the question.
The position of the object from the lens is given as $u = 1.5{\text{m}} = 1500{\text{mm}}$ and the focal length of the lens is given as $f = 60{\text{mm}}$.
The distance between the film and the lens, i.e., the image distance $v$ is unknown.
Step 2: Using the thin lens formula find the image distance.
Let $v'$ be the image distance for the nearby object. Then the thin lens formula will be
$\dfrac{1}{{v'}} - \dfrac{1}{u} = \dfrac{1}{f}$ ---------- (1)
where $v'$ is the position of the image from the lens, $u$ is the position of the nearby object from the lens and $f$ is the focal length of the lens.
Substituting the values for $u = - 1500{\text{mm}}$ and $f = 60{\text{mm}}$ in equation (1) we get, $\dfrac{1}{{v'}} + \dfrac{1}{{1500}} = \dfrac{1}{{60}}$ or on rearranging we get, $\dfrac{1}{{v'}} = \dfrac{1}{{60}} - \dfrac{1}{{1500}}$
Calculating the right-hand side of the above equation we get, we get, $\dfrac{1}{{v'}} = \dfrac{{25 - 1}}{{1500}} = \dfrac{{24}}{{1500}}$
Taking the reciprocal we get, $v' = \dfrac{{1500}}{{24}} = 62.5{\text{mm}}$
Thus the position of the image of the nearby object is $v' = 62.5{\text{mm}}$ .
Step 3: Find the change in the distance of the image formed when the focus was shifted from a distant object to a nearby object.
The distant object is assumed to be at infinity. So, for a convex lens if the object is at infinity then the image formed will be at the principal focus of the lens. This implies that the image distance for the distant object is $v = f = 60{\text{mm}}$.
Then the change in distance between the film and the lens will be $\Delta v = v' - v$ .
Substituting for $v = 60{\text{mm}}$ and $v' = 62.5{\text{mm}}$ in the above relation we get, $\Delta v = 62.5 - 60 = + 2.5{\text{mm}}$
Thus the distance between the film and the lens is increased by 2.5 mm.
Hence the correct option is B.

Note: By sign convention, the distance of the object from the convex lens $u$ is always negative while the focal length of the convex lens $f$ is positive. For the camera lens, the focal length is fixed. As the object distance changes the distance between the lens and the film adjusts so that it matches the image distance.