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Q. A telescope of aperture $3 \times 10^{-2}$ m diameter is focused on a window at $80 \,m$ distance fitted with a wire mesh of spacing $3 \times 10^{-3}$ m. Given: $\lambda = $ $5.5 \times 10^{-7}$ m, which of the following is true for observing the mesh through the telescope?

AIEEEAIEEE 2012Ray Optics and Optical Instruments

Solution:

Given : $d= 3 \times 10^{-2}\, m$
$\lambda = 5.5 \times 10^{-7}\,m$
Limit of resolution, $\Delta\theta = \frac{1.22\lambda}{d}$
$= \frac{1.22\times 5.5\times 10^{-7}}{3\times 10^{-2}} = 2.23 \times 10^{-5}\,rad.$
At a distance of 80 m , the telescope is able to resolve between two points which are separated by $2.23 \times 10^{-5}\times80\,m$
$= 1.78 \times 10^{-3} \,m$