Focal Length in Displacement Method – Rankers Physics
Topic: Ray Optics
Subtopic: Lens Makers Formula/ Len's Formula

Focal Length 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, the magnification values are \(m_1 = \frac{v_1}{u_1}\) and \(m_2 = \frac{v_2}{u_2} = \frac{u_1}{v_1}\). Also, \(x = v_1 - u_1\). Thus, \(m_1 - m_2 = \frac{x}{f}\), which gives \(f = \frac{x}{m_1 - m_2}\).

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