Focal length of lens in displacement method – Rankers Physics
Topic: Ray Optics
Subtopic: Lens Makers Formula/ Len's Formula

Focal length of lens in displacement method

In the displacement method, a convex lens is placed in between an object and a screen. If magnification in the two positions are \(m_1\) and \(m_2\) (\(m_1 > m_2\)) and the distance between two positions of the lens is x, the focal length of the lens is
\(\frac{x}{m_1+m_2}\)
\(\frac{x}{m_1-m_2}\)
\(\frac{x}{(m_1+m_2)^2}\)
\(\frac{x}{(m_1-m_2)^2}\)

Solution:

In displacement method, \(m_1 = \frac{v_1}{u_1}\) and \(m_2 = \frac{v_2}{u_2} = \frac{u_1}{v_1}\). Since the distance between two positions is \(x = v_1 - u_1\), we obtain \(m_1 - m_2 = \frac{x}{f}\), hence \(f = \frac{x}{m_1-m_2}\).

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