Answer to Question #115941 in Optics for Shavy

Question

In the displacement method,a convex lens is placed in between an object and a screen. If the magnification in the two positions are m1 and m2 and the displacement of the lens between the two positions is x,then the focal length of the lens is
(1)x/m1+m2 (2)x/m1-m2 (3) x(m1+m2)² (4)x/(m1-m2)²

Answer 1

For magnification $m_1$:

$$i_1=-m_1o_1,\\ \frac{1}{f}=\frac{1}{o_1}+\frac{1}{i_1}=\frac{1}{o_1}-\frac{1}{m_1o_1}=\frac{1}{o_1}(1-1/m_1).$$

By analogy for magnification $m_2$:

$$\frac{1}{f}=\frac{1}{o_2}(1-1/m_2)$$

From the figure we see that

$$x=o_1-o_2\rightarrow o_2=o_1-x,$$

substitute this:

$$\frac{1}{f}=\frac{1}{o_1-x}(1-1/m_2),$$

solve the system that consists of the last equation and this one:

$$\frac{1}{f}=\frac{1}{o_1}(1-1/m_1).$$

Simple math gives us the solution:

$$f=\frac{x}{m_1-m_2}.$$