Saved Bookmarks
| 1. |
A microscope has objective of aperture 8 mm and focal length 2.5 cm. Estimate its resolving power. Given lamda=5500Å |
Answer» Solution :We assume that the object is placed a little beyond its focal distance, say 2.5 cm. `tan alpha = (4mm)/(2.5cm) = (0.4)/(2.5) = 0.16` Since `SIN alpha` small , `sin alpha ~~ tan alpha = 0.16` Thus ,` Delta x = (1.22 lamda)/(2mu sin alpha) = (1.22 (5500 xx 10^(-10) m))/(2 xx 1 xx 0.16) = 2 xx 10^(-6) cm` |
|