Question

A lens placed $10\,cm$ away from a wall casts a sharp inverted image of a candle on it. It again casts a sharp image when the lens is moved $20\,cm$ further away from the wall. Now. the candle and the lens are moved such that a sharp inverted image with unit magnification is formed on the wall. To achieve this configuration, the candle was moved

• A. $20\,cm$ towards the wall.
• B. $20\,cm$ away from the wall.
• C. $10\,cm$ away from the
• D. $10\,cm$ towards the wall.

Answer 1

Answer: (D)

10 cm towards the wall

Answer 2

To solve this problem, we first need to determine the focal length of the lens and the initial distance between the candle and the wall. Then, we will use this information to find the new positions for unit magnification and calculate the required movement of the candle.

Part 1: Determine the focal length ($f$) and the initial object-screen distance ($D$)

Let the distance between the candle (object) and the wall (screen) be $D$. We use the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. For a real, inverted image formed by a converging lens, $u$ is negative and $v$ is positive. Let $|u|$ be the magnitude of the object distance. So, $u = -|u|$. The lens formula becomes: $\frac{1}{v} + \frac{1}{|u|} = \frac{1}{f}$. Also, the object-screen distance is $D = |u| + v$. Therefore, $|u| = D - v$. Substituting $|u|$ into the lens formula: $\frac{1}{v} + \frac{1}{D-v} = \frac{1}{f}$ $\frac{D-v+v}{v(D-v)} = \frac{1}{f}$ $\frac{D}{v(D-v)} = \frac{1}{f}$ So, $f = \frac{v(D-v)}{D}$.

Scenario 1: The lens is placed $10\,cm$ away from the wall. So, the image distance $v_1 = +10\,cm$. From the formula derived: $f = \frac{10(D-10)}{D}$ (Equation 1)

Scenario 2: The lens is moved $20\,cm$ further away from the wall. This means the new position of the lens is $10 + 20 = 30\,cm$ away from the wall. So, the image distance $v_2 = +30\,cm$. The object-screen distance $D$ remains constant. From the formula derived: $f = \frac{30(D-30)}{D}$ (Equation 2)

Equating Equation 1 and Equation 2: $\frac{10(D-10)}{D} = \frac{30(D-30)}{D}$ Since $D \ne 0$, we can cancel $D$ from both sides: $10(D-10) = 30(D-30)$ Divide by 10: $D-10 = 3(D-30)$ $D-10 = 3D-90$ $2D = 80$ $D = 40\,cm$

Now substitute $D=40\,cm$ into Equation 1 to find $f$: $f = \frac{10(40-10)}{40} = \frac{10 \times 30}{40} = \frac{300}{40} = \frac{30}{4} = 7.5\,cm$.

So, the focal length of the lens is $7.5\,cm$, and the initial distance between the candle and the wall is $40\,cm$.

Part 2: Achieve unit magnification

For a real, inverted image with unit magnification ($m = -1$), the object must be placed at $2f$ from the lens, and the image will be formed at $2f$ on the other side. So, $|u| = 2f$ and $v = 2f$. Using $f = 7.5\,cm$: $|u| = 2 \times 7.5\,cm = 15\,cm$. $v = 2 \times 7.5\,cm = 15\,cm$.

Since the image is formed on the wall, the lens must be $15\,cm$ away from the wall. The candle (object) must be $15\,cm$ away from the lens. Therefore, the new distance of the candle from the wall is $|u| + v = 15\,cm + 15\,cm = 30\,cm$.

Part 3: Calculate the movement of the candle

Initial distance of the candle from the wall = $D = 40\,cm$. Final distance of the candle from the wall = $30\,cm$.

The change in distance of the candle from the wall is $40\,cm - 30\,cm = 10\,cm$. Since the final distance is less than the initial distance, the candle was moved closer to the wall.

Therefore, the candle was moved $10\,cm$ towards the wall.