An object is located in a fixed position in front of a screen. Sharp image is obtained on the screen for two positions of a thin lens separated by 10 cm. The size of the images in two situations are in the ratio 3 : 3. What is the distance between the screen and the object?

In the displacement method, an object and screen are fixed at distance $$D$$ apart, and a thin lens forms a sharp image at two positions separated by $$d = 10$$ cm. For the two lens positions, the object and image distances are $$u_1 = \dfrac{D - d}{2}$$, $$v_1 = \dfrac{D + d}{2}$$ and $$u_2 = \dfrac{D + d}{2}$$, $$v_2 = \dfrac{D - d}{2}$$.

The linear magnifications are $$m_1 = \dfrac{v_1}{u_1} = \dfrac{D + d}{D - d}$$ and $$m_2 = \dfrac{v_2}{u_2} = \dfrac{D - d}{D + d}$$. The ratio of the image areas is $$\dfrac{m_1^2}{m_2^2} = \left(\dfrac{D+d}{D-d}\right)^4$$. Given that this ratio equals $$\dfrac{3}{2}$$, we get $$\left(\dfrac{D+d}{D-d}\right)^2 = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$$.

Taking the square root: $$\dfrac{D+d}{D-d} = \left(\dfrac{3}{2}\right)^{1/4}$$. However, a cleaner interpretation is that the linear magnification ratio satisfies $$\dfrac{m_1}{m_2} = \left(\dfrac{D+d}{D-d}\right)^2 = \dfrac{3}{2}$$, giving $$\dfrac{D+d}{D-d} = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$$.

Cross-multiplying: $$2(D + d) = \sqrt{6}(D - d)$$, so $$2D + 2d = \sqrt{6}\,D - \sqrt{6}\,d$$, which rearranges to $$D(\sqrt{6} - 2) = d(2 + \sqrt{6})$$. With $$d = 10$$: $$D = \dfrac{10(2 + \sqrt{6})}{\sqrt{6} - 2}$$.

Rationalizing by multiplying numerator and denominator by $$(\sqrt{6} + 2)$$: $$D = \dfrac{10(2 + \sqrt{6})(\sqrt{6} + 2)}{6 - 4} = \dfrac{10(2\sqrt{6} + 4 + 6 + 2\sqrt{6})}{2} = \dfrac{10(10 + 4\sqrt{6})}{2} = 50 + 20\sqrt{6}$$.

Since $$\sqrt{6} \approx 2.449$$, we get $$D = 50 + 48.99 \approx 99.0$$ cm.