Q69 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500$ nm, at a resolution of $5 \times 10^{-7}$ radian. The value of $R$ is _____ cm.
- A$61$
- Correct answerB$122$
- C$244$
- D$305$
Solution
The limit of resolution of a telescope is given by the formula:
$\Delta \theta = \dfrac{1.22 \lambda}{R}$
where $\lambda$ is the wavelength of light and $R$ is the diameter of the objective lens.
Given:
$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$
$\Delta \theta = 5 \times 10^{-7} \text{ rad}$
Substituting the values into the formula:
$5 \times 10^{-7} = \dfrac{1.22 \times 5 \times 10^{-7}}{R}$
$R = 1.22 \text{ m}$
Converting to centimeters:
$R = 1.22 \times 100 \text{ cm} = 122 \text{ cm}$
Answer: $122$
$\Delta \theta = \dfrac{1.22 \lambda}{R}$
where $\lambda$ is the wavelength of light and $R$ is the diameter of the objective lens.
Given:
$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$
$\Delta \theta = 5 \times 10^{-7} \text{ rad}$
Substituting the values into the formula:
$5 \times 10^{-7} = \dfrac{1.22 \times 5 \times 10^{-7}}{R}$
$R = 1.22 \text{ m}$
Converting to centimeters:
$R = 1.22 \times 100 \text{ cm} = 122 \text{ cm}$
Answer: $122$
Hence, the correct answer is option: 'B'.