(B) The angular resolution of a telescope is given by the formula $\theta = 1.22 \frac{\lambda}{R}$.Given:$\theta = 5 \times 10^{-7} \text{ rad}$$\lam

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(B) The angular resolution of a telescope is given by the formula $\theta = 1.22 \frac{\lambda}{R}$.Given:$\theta = 5 \times 10^{-7} \text{ rad}$$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$Substituting these values into the formula:$5 \times 10^{-7} = 1.22 \times \frac{500 \times 10^{-9}}{R}$Rearranging for $R$:$R = \frac{1.22 \times 500 \times 10^{-9}}{5 \times 10^{-7}}$$R = 1.22 \times 100 \times 10^{-2}$$R = 1.22 \text{ m}$Converting to centimeters:$R = 1.22 \times 100 \text{ cm} = 122 \text{ cm}$.

Class 12 Physics Wave Optics Resolving power-Human eye, Microscopes and Telescopes

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$A$ telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500 \text{ nm}$, at a resolution of $5 \times 10^{-7} \text{ radian}$. The value of $R$ is . . . . . . $\text{cm}$.

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Wave Optics

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Resolving power-Human eye, Microscopes and Telescopes

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The wavelengths of light used in an optical instrument are ${\lambda _1 = 4000 \ \text{\AA}}$ and ${\lambda _2 = 5000 \ \text{\AA}}$. The ratio of their respective resolving powers (corresponding to ${\lambda _1}$ and ${\lambda _2}$) is:

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The diameter of the objective of a telescope is $3.6 \ m$. The limit of resolution of the telescope for a light of wavelength $540 \ nm$ is

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The aperture of the objective of a telescope is $24.4 \, cm$. What is the resolving power of this telescope if light of wavelength $2440 \, \mathring{A}$ is used to view the object?

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$A$ microscope has an objective of aperture $8 \text{ mm}$ and focal length of $5 \text{ cm}$. The minimum separation between two objects to be just resolved by the microscope is (wavelength of light used $= 5500 \text{ Å}$) (in $\mu\text{m}$)

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When the object is self-luminous,the resolving power of a microscope is given by the expression

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$A$ telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500 \text{ nm}$, at a resolution of $5 \times 10^{-7} \text{ radian}$. The value of $R$ is . . . . . . $\text{cm}$.

Similar Questions

The wavelengths of light used in an optical instrument are ${\lambda _1 = 4000 \ \text{\AA}}$ and ${\lambda _2 = 5000 \ \text{\AA}}$. The ratio of their respective resolving powers (corresponding to ${\lambda _1}$ and ${\lambda _2}$) is:

The diameter of the objective of a telescope is $3.6 \ m$. The limit of resolution of the telescope for a light of wavelength $540 \ nm$ is

The aperture of the objective of a telescope is $24.4 \, cm$. What is the resolving power of this telescope if light of wavelength $2440 \, \mathring{A}$ is used to view the object?

$A$ microscope has an objective of aperture $8 \text{ mm}$ and focal length of $5 \text{ cm}$. The minimum separation between two objects to be just resolved by the microscope is (wavelength of light used $= 5500 \text{ Å}$) (in $\mu\text{m}$)

When the object is self-luminous,the resolving power of a microscope is given by the expression

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