Solution to Question 22
(a) Finding the Focal Length
In the displacement method, the distance between the source and screen is $D = 120 \text{ cm}$. Let the focal length be $f$. If $I_1$ and $I_2$ are the sizes of the two sharp images formed at two positions of the lens, the magnifications are $m_1$ and $m_2$. We know that $m_1 m_2 = 1$. The ratio of image sizes is given as $1:9$. Let $\frac{I_1}{I_2} = \frac{1}{9}$. Since size $\propto |m|$, we have: $$\frac{m_1}{m_2} = \frac{1}{9} \quad \text{or} \quad \frac{m_1}{m_2} = 9$$ From the property of conjugate positions in displacement method, if $m_1 = \frac{v}{u}$, then $m_2 = \frac{u}{v}$. Therefore: $$\frac{m_1}{m_2} = \frac{v/u}{u/v} = \frac{v^2}{u^2} = 9 \implies \frac{v}{u} = 3$$ (Taking the case where $v > u$. The other case gives $1/3$, representing the same physical setup swapped). We have the system of equations: 1. $\frac{v}{u} = 3 \implies v = 3u$ 2. $v + u = D = 120 \text{ cm}$ Substituting (1) into (2): $$3u + u = 120 \implies 4u = 120 \implies u = 30 \text{ cm}$$ $$v = 3(30) = 90 \text{ cm}$$ Now, using the lens formula to find $f$: $$f = \frac{uv}{u+v} = \frac{30 \times 90}{120} = \frac{2700}{120} = 22.5 \text{ cm}$$
(b) Ratio of Brightness
Brightness (Intensity) of an image is defined as the power per unit area. The power collected by the lens depends on the solid angle subtended by the lens aperture at the source. $$P_{collected} \propto \frac{Area_{lens}}{u^2} \propto \frac{1}{u^2}$$ The area of the image is proportional to the square of the magnification $m$. $$Area_{image} \propto m^2 = \left(\frac{v}{u}\right)^2$$ Therefore, the brightness $B$ is: $$B \propto \frac{P_{collected}}{Area_{image}} \propto \frac{1/u^2}{v^2/u^2} = \frac{1}{v^2}$$ We compare the brightness of the two images formed at distances $v_1$ and $v_2$.
- Case 1 (Smaller image): $m = 1/3$. Here $v_1 = 30 \text{ cm}, u_1 = 90 \text{ cm}$.
- Case 2 (Larger image): $m = 3$. Here $v_2 = 90 \text{ cm}, u_2 = 30 \text{ cm}$.