Question

A converging lens (focal length 'f' ) is broken in two equal pieces and placed at 60 cm as shown along with the object. It is found that real images are formed at the same place and ratio of image heights is 9 : 1 , if the value of 'f'

• A. 22.5

Answer 1

Answer: (A)

22.5

Answer 2

Let the object be placed at a distance $u$ from the first lens piece. Let the first lens piece be at $x=0$ and the second lens piece be at $x=60$ cm. The object is at $x=-u$. The focal length of each piece is $f$.

For the first lens piece at $x=0$: Object distance $u_1 = -u$. Let the image distance be $v_1$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f}$: $\frac{1}{v_1} - \frac{1}{-u} = \frac{1}{f} \implies \frac{1}{v_1} + \frac{1}{u} = \frac{1}{f}$ $v_1 = \frac{fu}{u-f}$. For a real image, $v_1 > 0$, which requires $u > f$ (since $f>0, u>0$). The image is formed at $x = v_1$.

For the second lens piece at $x=60$: The object is at $x=-u$. The distance of the object from the second lens piece is $60 - (-u) = 60+u$. Object distance $u_2 = -(60+u)$ (since the object is to the left of the lens). Let the image distance be $v_2$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f}$: $\frac{1}{v_2} - \frac{1}{-(60+u)} = \frac{1}{f} \implies \frac{1}{v_2} + \frac{1}{60+u} = \frac{1}{f}$ $v_2 = \frac{f(60+u)}{60+u-f}$. For a real image, $v_2 > 0$, which requires $60+u > f$. The image formed by the second piece is at $x = 60+v_2$.

The real images are formed at the same place, so the x-coordinates of the images are equal: $v_1 = 60 + v_2$. Substitute the expressions for $v_1$ and $v_2$: $\frac{fu}{u-f} = 60 + \frac{f(60+u)}{60+u-f}$ $\frac{fu}{u-f} - \frac{f(60+u)}{60+u-f} = 60$ $\frac{fu(60+u-f) - f(60+u)(u-f)}{(u-f)(60+u-f)} = 60$ The numerator is $f [u(60+u-f) - (60+u)(u-f)] = f [ (60u+u^2-uf) - (60u-60f+u^2-uf) ] = f [ 60u+u^2-uf - 60u+60f-u^2+uf ] = f(60f) = 60f^2$. So, $\frac{60f^2}{(u-f)(60+u-f)} = 60$ $f^2 = (u-f)(60+u-f)$ $f^2 = 60u + u^2 - uf - 60f - uf + f^2$ $0 = u^2 + (60-2f)u - 60f$.

The magnification for the first piece is $m_1 = \frac{v_1}{u_1} = \frac{v_1}{-u}$. The magnification for the second piece is $m_2 = \frac{v_2}{u_2} = \frac{v_2}{-(60+u)}$. The height of the image formed by the first piece is $h_1 = |m_1| h_o = |\frac{v_1}{-u}| h_o = \frac{v_1}{u} h_o$ (since $v_1 > 0, u > 0$). The height of the image formed by the second piece is $h_2 = |m_2| h_o = |\frac{v_2}{-(60+u)}| h_o = \frac{v_2}{60+u} h_o$ (since $v_2 > 0, 60+u > 0$).

The ratio of image heights is given as 9:1. So $\frac{h_1}{h_2} = 9$ or $\frac{h_2}{h_1} = 9$. Case 1: $\frac{h_1}{h_2} = 9$ $\frac{v_1/u}{v_2/(60+u)} = 9 \implies \frac{v_1(60+u)}{v_2 u} = 9$. Substitute $v_1 = \frac{fu}{u-f}$ and $v_2 = \frac{f(60+u)}{60+u-f}$: $\frac{\frac{fu}{u-f} (60+u)}{\frac{f(60+u)}{60+u-f} u} = 9$ $\frac{fu(60+u)(60+u-f)}{f(60+u)u(u-f)} = 9$ $\frac{60+u-f}{u-f} = 9$ $60+u-f = 9(u-f)$ $60+u-f = 9u - 9f$ $60 + 8f = 8u$ $u - f = \frac{60}{8} = 7.5$. So $u = f + 7.5$.

Substitute $u = f + 7.5$ into the quadratic equation $u^2 + (60-2f)u - 60f = 0$: $(f+7.5)^2 + (60-2f)(f+7.5) - 60f = 0$ $(f^2 + 15f + 56.25) + (60f + 450 - 2f^2 - 15f) - 60f = 0$ $f^2 + 15f + 56.25 + 60f + 450 - 2f^2 - 15f - 60f = 0$ $(f^2 - 2f^2) + (15f + 60f - 15f - 60f) + (56.25 + 450) = 0$ $-f^2 + 0f + 506.25 = 0$ $f^2 = 506.25$ $f = \sqrt{506.25} = 22.5$ cm.

We should check if this value of $f$ is consistent with the real image conditions $u>f$ and $60+u>f$. If $f=22.5$, then $u = f+7.5 = 22.5+7.5 = 30$. $u=30 > f=22.5$, so the first condition is satisfied. $60+u = 60+30 = 90$. $90 > f=22.5$, so the second condition is satisfied. Both images are real.

Case 2: $\frac{h_2}{h_1} = 9$ $\frac{v_2/(60+u)}{v_1/u} = 9 \implies \frac{v_2 u}{v_1(60+u)} = 9$. Using the same substitution as before: $\frac{\frac{f(60+u)}{60+u-f} u}{\frac{fu}{u-f} (60+u)} = 9$ $\frac{f(60+u)u(u-f)}{fu(60+u)(60+u-f)} = 9$ $\frac{u-f}{60+u-f} = 9$ $u-f = 9(60+u-f)$ $u-f = 540 + 9u - 9f$ $8f - 8u = 540$ $f - u = \frac{540}{8} = 67.5$. So $u = f - 67.5$. Since $u$ is the magnitude of the object distance, $u>0$. This implies $f > 67.5$. Substitute $u = f - 67.5$ into the quadratic equation $u^2 + (60-2f)u - 60f = 0$: $(f-67.5)^2 + (60-2f)(f-67.5) - 60f = 0$ $(f^2 - 135f + 67.5^2) + (60f - 60 \times 67.5 - 2f^2 + 135f) - 60f = 0$ $(f^2 - 135f + 4556.25) + (60f - 4050 - 2f^2 + 135f) - 60f = 0$ $(f^2 - 2f^2) + (-135f + 60f + 135f - 60f) + (4556.25 - 4050) = 0$ $-f^2 + 0f + 506.25 = 0$ $f^2 = 506.25$ $f = \sqrt{506.25} = 22.5$ cm.

In this case, $f=22.5$ cm. Then $u = f - 67.5 = 22.5 - 67.5 = -45$. The magnitude of the object distance $u$ must be positive. The initial setup assumed the object is to the left of the first lens piece, so its coordinate is $-u$ where $u>0$. If $u=-45$, it means the object is at $x=45$. Let's reconsider the setup with the object at $x=45$. The first lens is at $x=0$, the second lens is at $x=60$. Object at $x=45$. For the first lens at $x=0$: Object distance $u_1 = 45-0 = 45$. This is to the right, so $u_1 = +45$. $\frac{1}{v_1} - \frac{1}{45} = \frac{1}{22.5} = \frac{1}{45/2} = \frac{2}{45}$. $\frac{1}{v_1} = \frac{2}{45} + \frac{1}{45} = \frac{3}{45} = \frac{1}{15}$. So $v_1 = 15$. Image is at $x=0+15=15$. This is a real image. Magnification $m_1 = \frac{v_1}{u_1} = \frac{15}{45} = \frac{1}{3}$. Height $h_1 = \frac{1}{3} h_o$.

For the second lens at $x=60$: Object at $x=45$. Distance is $60-45=15$. This is to the left of the second lens, so $u_2 = -15$. $\frac{1}{v_2} - \frac{1}{-15} = \frac{1}{22.5} = \frac{2}{45}$. $\frac{1}{v_2} + \frac{1}{15} = \frac{2}{45}$. $\frac{1}{v_2} = \frac{2}{45} - \frac{1}{15} = \frac{2-3}{45} = -\frac{1}{45}$. So $v_2 = -45$. Image is at $x=60+(-45)=15$. This is a virtual image. The problem states that real images are formed at the same place. So this case ($u=f-67.5$) is not valid because it leads to a virtual image from the second lens.

The only valid case is $u = f + 7.5$, which gives $f=22.5$ cm and $u=30$ cm. Let's check the image heights ratio for this case. $u=30, f=22.5$. $v_1 = \frac{fu}{u-f} = \frac{22.5 \times 30}{30 - 22.5} = \frac{22.5 \times 30}{7.5} = 3 \times 30 = 90$. Image at $x=90$. $v_2 = \frac{f(60+u)}{60+u-f} = \frac{22.5 \times (60+30)}{60+30 - 22.5} = \frac{22.5 \times 90}{90 - 22.5} = \frac{22.5 \times 90}{67.5} = \frac{22.5}{67.5} \times 90 = \frac{1}{3} \times 90 = 30$. Image at $x=60+30=90$. The images are indeed formed at the same place ($x=90$). Both are real ($v_1>0, v_2>0$).

Magnification $m_1 = \frac{v_1}{u_1} = \frac{90}{-30} = -3$. $h_1 = |m_1| h_o = 3 h_o$. Magnification $m_2 = \frac{v_2}{u_2} = \frac{30}{-(60+30)} = \frac{30}{-90} = -\frac{1}{3}$. $h_2 = |m_2| h_o = \frac{1}{3} h_o$. The ratio of image heights is $\frac{h_1}{h_2} = \frac{3 h_o}{(1/3) h_o} = 9$. This matches the given ratio 9:1.

Therefore, the value of $f$ is 22.5 cm.

The final answer is $\boxed{22.5}$.