Is focal length of convex lens negative when it is surrounded by a medium with greater refractive index than that of lens?
[Physics] Can focal length of convex lens be negative
lensesopticsrefraction
Related Solutions
A lens has two surfaces. One of these surfaces still has refracting power, due to the difference in refractive index and angle of incidence.
Refraction on one surface
Let parallel rays start in glas. Radius sign convention results $R=-200\,mm$ since incident light is on same side as center of curvature.
Radius of curvature and the two media influence the focal point. For your rays propagate from right to left (air) the FFL (front focal length, as depicted) is $$f=R\cdot \frac{n_1}{n_1-n_2}=R\cdot \frac{1}{1-n_2}$$ This is the reciprocal element of your transfer matrix. Focal length is proportional to radius of curvature. Increasing refractive index $n_2$ of glas shortens it. $$f=-200\,mm \cdot \frac{1}{-0.5}=400\,mm$$
Remark
The cited above lensmaker equation with $R_1=\infty$ will return the same result. Since on-axis parallel rays are not refracted on plane surface, as can be seen from Snell's law.
The sign of the focal length will change. So, a concave lens would behave like a convex lens, and a convex lens as concave lens.
The lens maker’s formula for a thin lens is: $$\dfrac{1}{f} = \left(\dfrac{\mu_2}{\mu_1} -1\right)\left(\dfrac{1}{R _1} - \dfrac{1}{R_2}\right)$$
Where $\mu_2$ is the refractive index of the lens, and $\mu_1$ is the refractive index of the medium.
If $\mu_1 > \mu_2$, then $\frac{\mu_2}{\mu_1} -1 < 0$.
So, the focal length of a convex lens will become negative, and it will act like a concave lens, and focal lens of a concave lens will become positive, so it'll act as a convex lens.
Best Answer
A convex lens in a medium with index larger than that of the lens will act like a divergent lens, yes. This is the physical fact. If you consider the focal length positive or negative will depend on the convention used. By usual convention, the focal lens of divergent lenses is taken with minus sign. If you use the lens-maker formula (in the thin lens approximation) $$ \frac{1}{f}=(n_{lens} -n_{medium})(\frac{1}{R_1}- \frac{1}{R_2}) $$ you can see that if $ \ n_{lens} <n_{medium}$ the focal length will be negative. Note that the formula is written with the convention that $R_2$ is negative for convex lenses so that the content of the second parenthesis on the RHS is always positive for convex lenses.