Question

Hubble telescope uses mean wavelength $\lambda =550nm.$ Aperture of objective is $12.2m$. It is established at a distance $570km$ from Earth. Find the smallest size of crater that can be identified in image by telescope.

• A. $24.46mm$
• B. $31.46mm$
• C. $20.46mm$
• D. $50.46mm$

Answer 1

The correct option is B $31.46mm$


Resolution Power =$\frac{A}{1.22\lambda }=\frac{12.2}{1.22\times 550\times {10}^{-9}}=\frac{{10}^{10}}{550}=\frac{{10}^9}{55}R.P=\frac{1}{{\theta }_{min}}=\frac{{10}^9}{55}$
$\theta =\frac{X}{570\times {10}^3}$
For clear image,
$\theta \ge {\theta }_{min}\frac{X}{570\times {10}^3}\ge \frac{55}{{10}^9}X\ge 31.46mm$