Question

In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images formed by a convex lens placed between them is 4.84. Choose correct statement(s).

• A. Ratio of length of object to the length of shorter image is $\frac{11}{5}$
• B. Distance between the two positions of the lens is 36 cm
• C. Focal length of the lens is 22.5 cm
• D. Distance of the lens from the shorter image is 30 cm

Answer 1

(A), (B), (D)

Answer 2

In the displacement method, let the distance between the object and the screen be $D$. Let the two positions of the convex lens between the object and the screen produce real images on the screen. Let the distances of the lens from the object in the two positions be $u_1$ and $u_2$, and the corresponding distances of the lens from the screen (image distances) be $v_1$ and $v_2$.

For both positions, we have $u+v=D$. The magnifications are $m_1 = \frac{v_1}{u_1}$ and $m_2 = \frac{v_2}{u_2}$. The lengths of the images are $I_1 = |m_1|O$ and $I_2 = |m_2|O$, where $O$ is the length of the object.

In the displacement method, the two positions are conjugate, meaning $u_1 = v_2$ and $u_2 = v_1$. The distance between the two lens positions is $x = |u_1 - u_2|$.

We also have $u_1 + v_1 = D$ and $u_2 + v_2 = D$. Since $u_1 = v_2$ and $u_2 = v_1$, we have $u_1 + u_2 = D$.

The magnifications are $|m_1| = \frac{v_1}{u_1}$ and $|m_2| = \frac{v_2}{u_2}$. The product of the magnitudes of magnifications is $|m_1 m_2| = \frac{v_1}{u_1} \frac{v_2}{u_2} = \frac{v_1}{u_1} \frac{u_1}{v_1} = 1$.

The ratio of the lengths of the two images is given as $\frac{I_1}{I_2} = \frac{|m_1|O}{|m_2|O} = \frac{|m_1|}{|m_2|} = 4.84$.

Since $|m_1 m_2| = 1$, we have $|m_1| = \sqrt{4.84} = 2.2$ and $|m_2| = \frac{1}{2.2} = \frac{10}{22} = \frac{5}{11}$.

Let $I_1$ be the length of the larger image and $I_2$ be the length of the shorter image. Then $\frac{I_1}{I_2} = \frac{|m_1|}{|m_2|} = 4.84$.

So, the magnifications are $m_{large} = 2.2 = \frac{11}{5}$ and $m_{small} = \frac{5}{11}$.

Let's find the object and image distances for each case.

For the larger image ($m = \frac{11}{5}$): $\frac{v}{u} = \frac{11}{5}$, so $v = \frac{11}{5}u$. Also $u+v = D = 96$.

$u + \frac{11}{5}u = 96 \implies \frac{16}{5}u = 96 \implies u = 96 \times \frac{5}{16} = 6 \times 5 = 30$ cm.

$v = 96 - u = 96 - 30 = 66$ cm.

This corresponds to one position of the lens: $u_1 = 30$ cm, $v_1 = 66$ cm.

For the shorter image ($m = \frac{5}{11}$): $\frac{v}{u} = \frac{5}{11}$, so $v = \frac{5}{11}u$. Also $u+v = D = 96$.

$u + \frac{5}{11}u = 96 \implies \frac{16}{11}u = 96 \implies u = 96 \times \frac{11}{16} = 6 \times 11 = 66$ cm.

$v = 96 - u = 96 - 66 = 30$ cm.

This corresponds to the other position of the lens: $u_2 = 66$ cm, $v_2 = 30$ cm.

Note that $u_1 = v_2 = 30$ cm and $u_2 = v_1 = 66$ cm (magnitudes), confirming the displacement method property.

Now let's evaluate the statements:

(A) Ratio of length of object to the length of shorter image is $\frac{O}{I_{shorter}}$.

The shorter image has magnification $m_{small} = \frac{I_{shorter}}{O} = \frac{5}{11}$.

So, $\frac{O}{I_{shorter}} = \frac{1}{m_{small}} = \frac{11}{5}$.

Statement (A) is correct.

(B) Distance between the two positions of the lens is 36 cm.

The two positions of the lens are at distances $u_1 = 30$ cm and $u_2 = 66$ cm from the object.

The distance between the two positions is $x = |u_1 - u_2| = |30 - 66| = |-36| = 36$ cm.

Statement (B) is correct.

(C) Focal length of the lens is 22.5 cm.

Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (with sign convention, $u$ is negative).

Using magnitudes: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.

Using the first position ($u_1=30$, $v_1=66$): $\frac{1}{f} = \frac{1}{66} + \frac{1}{30} = \frac{5 + 11}{330} = \frac{16}{330}$.

$f = \frac{330}{16} = \frac{165}{8} = 20.625$ cm.

Using the second position ($u_2=66$, $v_2=30$): $\frac{1}{f} = \frac{1}{30} + \frac{1}{66} = \frac{11 + 5}{330} = \frac{16}{330}$.

$f = \frac{330}{16} = \frac{165}{8} = 20.625$ cm.

The focal length is 20.625 cm, not 22.5 cm.

Statement (C) is incorrect.

(D) Distance of the lens from the shorter image is 30 cm.

The shorter image is formed when the lens is at the position where the object distance is $u_2 = 66$ cm and the image distance is $v_2 = 30$ cm.

The distance of the lens from the shorter image (which is on the screen) is the image distance $v_2$.

$v_2 = 30$ cm.

Statement (D) is correct.

The correct statements are (A), (B), and (D).