What happens to a 50 ohm line when its terminated at an impedance less than 50 ohms?
For a quarter wavelength line you get the following relationship: - 

\$Z_{IN} = \dfrac{Z_0^2}{Z_L}\$ where \$Z_L\$ is the load impedance.

If the characteristic impedance is 50 ohm and the load is 25 ohm (to make my brain number crunching easier), Zin will be 2500/25 = 100 ohms. If the load was 250 ohms, Zin would be 10 ohms. That's what you get when the transmission line is exactly one quarter of a wavelength. This is the math behind it: -



The "tan" parts become infinite at a quarter wavelength leaving you with the basic equation at the top. For non-quarter wavelength lines the "tan" part plays a significant and varied role.

The bigger picture over a full wavelength: -



An open circuit t-line will look capacitive when very short and at a quarter wave will look like a short circuit. As length increases it looks inductive until at half a wavelength it looks like an open circuit again.