Chapter 19, Problem 4STP

To determine

To choose: The correct option.

Answer to Problem 4STP

The correct option is (C), $D=1.5\times {10}^{-2}\text{m}\text{}$

Explanation of Solution

Given:

Two stars are at a distance of 3.1 light-years apart. They are at an equal distance of 6.2×10⁴ light-years from a telescope in earth. It is observed with a light of wavelength of 610 nm .

Formula used:

Using Rayleigh criterion, two objects at the limit of resolution are separated by a distance $x_{obj}$ which can be expressed as the function of the wavelength of light, $\lambda$ , distance between the observation point at the object, $L_{obj}$ and the diameter of the circular aperture of the telescope $D$ as,

  $x_{\text{obj}}=\frac{1.22\lambda L_{obj}}{D}$

Calculation:

The distance is given in light years. The conversion from light years to metre is given by,

1 light-year= $9.46\times {10}^{15}\text{}\text{m}$

Here, the wavelength of light is $\lambda =610\text{ nm}$ , distance between object is 3.1 light-years and the distance between the stars and the earth is 6.2×10⁴ light-years. Substituting these values, the smallest diameter of the aperture of the telescope is given by

  $\begin{array}{c}D=\frac{1.22\lambda L_{obj}}{x_{\text{obj}}}\\ =\frac{1.22\times 610\times {10}^{-9}\times 6.2\times {10}^4}{3.1}\\ =1488.4\times {10}^{-5}\text{}\text{}\text{m}\\ \simeq 1.5\times {10}^{-2}\text{m}\text{}\end{array}$

Conclusion:

Therefore, the correct option for the smallest diameter of the aperture of the telescope is (C), $D=1.5\times {10}^{-2}\text{m}\text{}$ .