Question

A thin convex lens forms an image on a screen. The height of the first image is 1.2 cm. Without changing the distance between object and screen, the lens is moved along the optical axis, and it is observed that the height of the second clear image is 0.3 cm. If the distance between the object and screen is 120 cm, the distance between the two positions of the lens for obtaining clear image on screen is

Answer 1

40 cm

Answer 2

The problem describes a scenario involving a convex lens and the lens displacement method (also known as the conjugate foci method). In this method, for a fixed distance D between the object and the screen, there are two positions of the lens that form a clear image on the screen.

Let:

• $h_o$ be the height of the object.
• $h_{i1}$ be the height of the first image.
• $h_{i2}$ be the height of the second image.
• D be the distance between the object and the screen.
• d be the distance between the two positions of the lens.

Key Concepts and Formulas:

1. Object Height: For the two positions of the lens forming real images, the height of the object ($h_o$) is the geometric mean of the two image heights ($h_{i1}$ and $h_{i2}$): $h_o = \sqrt{h_{i1} h_{i2}}$

2. Magnification: The magnification ($m$) for a lens is given by $m = \frac{\text{Image Height}}{\text{Object Height}}$. For the two positions, the magnifications are $m_1 = \frac{h_{i1}}{h_o}$ and $m_2 = \frac{h_{i2}}{h_o}$. Also, for the lens displacement method, the object and image distances for the two positions are related. If $u_1, v_1$ are for the first position and $u_2, v_2$ for the second, then $u_1 = v_2$ and $v_1 = u_2$. The object and image distances can be expressed in terms of D and d: $u_1 = \frac{D-d}{2}$, $v_1 = \frac{D+d}{2}$ (for the position giving the larger image) $u_2 = \frac{D+d}{2}$, $v_2 = \frac{D-d}{2}$ (for the position giving the smaller image) The corresponding magnifications are: $m_1 = \frac{v_1}{u_1} = \frac{D+d}{D-d}$ (for the larger image) $m_2 = \frac{v_2}{u_2} = \frac{D-d}{D+d}$ (for the smaller image)

Given Values:

• Height of the first image, $h_{i1} = 1.2$ cm
• Height of the second image, $h_{i2} = 0.3$ cm
• Distance between the object and screen, D = 120 cm

Step-by-step Solution:

1. Calculate the height of the object ($h_o$): $h_o = \sqrt{h_{i1} h_{i2}} = \sqrt{1.2 \text{ cm} \times 0.3 \text{ cm}} = \sqrt{0.36 \text{ cm}^2} = 0.6 \text{ cm}$

2. Calculate the magnification for one of the images: Let's use the first image, which is larger ($h_{i1} = 1.2$ cm). $m_1 = \frac{h_{i1}}{h_o} = \frac{1.2 \text{ cm}}{0.6 \text{ cm}} = 2$

3. Use the magnification formula in terms of D and d: Since $m_1 > 1$, this corresponds to the lens position closer to the object, where $u$ is smaller and $v$ is larger. $m_1 = \frac{D+d}{D-d}$ Substitute the values of $m_1$ and D: $2 = \frac{120 + d}{120 - d}$

4. Solve for d: $2(120 - d) = 120 + d$ $240 - 2d = 120 + d$ $240 - 120 = d + 2d$ $120 = 3d$ $d = \frac{120}{3}$ $d = 40 \text{ cm}$

The distance between the two positions of the lens is 40 cm.