Question #115941
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
1
Expert's answer
2020-05-15T16:10:12-0400


For magnification m1m_1:


i1=m1o1,1f=1o1+1i1=1o11m1o1=1o1(11/m1).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 m2m_2:


1f=1o2(11/m2)\frac{1}{f}=\frac{1}{o_2}(1-1/m_2)

From the figure we see that


x=o1o2o2=o1x,x=o_1-o_2\rightarrow o_2=o_1-x,

substitute this:


1f=1o1x(11/m2),\frac{1}{f}=\frac{1}{o_1-x}(1-1/m_2),

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


1f=1o1(11/m1).\frac{1}{f}=\frac{1}{o_1}(1-1/m_1).

Simple math gives us the solution:


f=xm1m2.f=\frac{x}{m_1-m_2}.


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