Question

In displacement method the ratio of image heights is 9 : 4 if distance between the object and screen is 60 cm, then distance between the two positions of lens in cm is :

Answer 1

12

Answer 2

Let $O$ be the height of the object. Let $I_1$ and $I_2$ be the heights of the two images formed on the screen in the displacement method. We are given that the ratio of image heights is $9:4$. Let's assume $I_1 > I_2$, so $\frac{I_1}{I_2} = \frac{9}{4}$. Let $m_1$ and $m_2$ be the magnifications for the two positions of the lens. The image height is given by $I = |m| O$. So, $I_1 = |m_1| O$ and $I_2 = |m_2| O$. The ratio of image heights is $\frac{I_1}{I_2} = \frac{|m_1| O}{|m_2| O} = \frac{|m_1|}{|m_2|}$. In the displacement method, the two positions of the lens are conjugate. If the object distance is $u_1$ and image distance is $v_1$ for the first position, then for the second position, the object distance is $u_2 = v_1$ and image distance is $v_2 = u_1$. The magnifications are $m_1 = \frac{v_1}{u_1}$ and $m_2 = \frac{v_2}{u_2} = \frac{u_1}{v_1}$. For real images formed by a converging lens, $u$ is negative and $v$ is positive. Let $u_1, v_1$ be the magnitudes of the distances. Then $m_1 = v_1/u_1$ and $m_2 = v_2/u_2 = u_1/v_1$. The product of the magnifications is $m_1 m_2 = \frac{v_1}{u_1} \times \frac{u_1}{v_1} = 1$. We have $\frac{|m_1|}{|m_2|} = \frac{9}{4}$. Since $m_1 m_2 = 1$, both $m_1$ and $m_2$ must be positive (as they are ratios of magnitudes of distances). So $\frac{m_1}{m_2} = \frac{9}{4}$. We have the system of equations:

1. $\frac{m_1}{m_2} = \frac{9}{4}$
2. $m_1 m_2 = 1$ From (2), $m_2 = \frac{1}{m_1}$. Substitute this into (1): $\frac{m_1}{1/m_1} = \frac{9}{4} \implies m_1^2 = \frac{9}{4}$. Since $I_1 > I_2$, we assume $m_1 > m_2$. From $m_1 m_2 = 1$, if $m_1 > m_2$, then $m_1 > 1$ and $m_2 < 1$. So, $m_1 = \sqrt{\frac{9}{4}} = \frac{3}{2}$. Then $m_2 = \frac{1}{m_1} = \frac{1}{3/2} = \frac{2}{3}$. Check: $\frac{m_1}{m_2} = \frac{3/2}{2/3} = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$. This is consistent.

The distance between the object and the screen is $D$. For a real image formed by a converging lens, the object distance $u$ and image distance $v$ are on opposite sides of the lens. The distance between the object and the screen is $D = u+v$, where $u$ and $v$ are the magnitudes of the distances. For the first position of the lens, the magnification is $m_1 = \frac{v_1}{u_1} = \frac{3}{2}$, so $v_1 = \frac{3}{2} u_1$. The distance between the object and screen is $D = u_1 + v_1 = u_1 + \frac{3}{2} u_1 = \frac{5}{2} u_1$. We are given $D = 60$ cm. $60 = \frac{5}{2} u_1 \implies u_1 = 60 \times \frac{2}{5} = 24$ cm. Then $v_1 = \frac{3}{2} u_1 = \frac{3}{2} \times 24 = 36$ cm.

For the second position of the lens, the object distance is $u_2 = v_1 = 36$ cm and the image distance is $v_2 = u_1 = 24$ cm. The magnification is $m_2 = \frac{v_2}{u_2} = \frac{24}{36} = \frac{2}{3}$, which is consistent.

In the displacement method, the distance between the two positions of the lens, $x$, is the difference between the two image distances (or object distances) from the lens. $x = |v_1 - v_2| = |36 - 24| = 12$ cm. Alternatively, $x = |u_2 - u_1| = |36 - 24| = 12$ cm.

The distance between the two positions of the lens is 12 cm.