Focal length of the given convex lens when it is placed in air is f = + 20 cm
Refractive index of the given medium with respect to air is aμm = 1.25
And aμg = 1.5
New focal length of the given convex lens when placed in a medium is f'
\(\frac 1f = (a_{\mu g} - 1) \left[\left(\frac 1{R_1}\right) + \left(\frac 1{R_2}\right)\right]\) ........(1)
\(\frac 1{f'} = (m_{\mu g} - 1) \left[\left(\frac 1{R_1}\right) + \left(\frac 1{R_2}\right)\right]\) ........(2)
Dividing (1) by (2), we get
\(\frac {f'}f = \frac{(a_{\mu g} - 1)}{(m _{\mu g} - 1)}\)
\(= \frac{(1.5 - 1)}{(1.2 - 1)}\)
\(= \frac{0.5}{0.2}\)
\(= \frac 52\)
\(= 2.5\)
\(f' = 2.5f\)
\(= (2.5 \times 20)cm\)
\(= +50 cm\)
As \(m_{\mu g} = \frac{\mu _g}{\mu _m}\)
\(= \frac{1.5}{1.25} \)
\(= 1.2\)
New focal length is positive.
The significance of the positive sign of the focal length is that the given convex lens is still converging in the given medium.
