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Question

Resolving power of a telescope can be increased by increasing:

The correct answer is

Diameter of Objective lens

Understanding Telescope Resolving Power

Resolving power of a telescope refers to its ability to distinguish between two objects that are very close together. A higher resolving power means the telescope can show finer details and separate objects that appear as a single blurry point with lower resolving power.

This capability is limited by the wave nature of light and the phenomenon of diffraction. When light from distant objects enters the telescope's objective lens (or mirror), it diffracts, creating a pattern rather than a sharp point image. The size of this diffraction pattern determines how close two points can be before their patterns overlap so much that they cannot be distinguished as separate.

Factors Affecting Telescope Resolving Power

The theoretical resolving power of a telescope is primarily determined by two factors:

  1. The wavelength ($\lambda$) of the light being observed.
  2. The diameter ($D$) of the telescope's objective lens or mirror (often called the aperture).

The minimum angular separation ($\Delta \theta$) between two point objects that a telescope can resolve is given by the Rayleigh criterion:

$$\Delta \theta = \frac{1.22 \lambda}{D}$$

where:

  • $\Delta \theta$ is the minimum resolvable angle in radians.
  • $\lambda$ is the wavelength of light.
  • $D$ is the diameter of the objective lens or mirror.

Resolving power is inversely proportional to the minimum resolvable angle ($\Delta \theta$). Therefore, resolving power is proportional to $\frac{1}{\Delta \theta}$.

So, resolving power $\propto \frac{1}{1.22 \lambda / D} \propto \frac{D}{\lambda}$.

This formula tells us that to increase the resolving power of a telescope, you should:

  • Increase the diameter ($D$) of the objective lens.
  • Decrease the wavelength ($\lambda$) of the light being observed.

Analyzing the Options for Increasing Resolving Power

Let's examine each option provided based on the formula for resolving power:

  1. Wavelength of light incident at telescope: If the wavelength ($\lambda$) increases, the minimum resolvable angle ($\Delta \theta$) increases ($\Delta \theta \propto \lambda$). Since resolving power is inversely proportional to $\Delta \theta$, increasing the wavelength decreases the resolving power. This option is incorrect.
  2. Diameter of eye piece: The eyepiece lens magnifies the image formed by the objective lens. While it affects the overall magnification you see, the resolving power is fundamentally set by the objective lens's diameter. A larger eyepiece diameter might provide a wider field of view or be more comfortable to look through, but it does not increase the resolving power. This option is incorrect.
  3. Focal length of eyepiece: Similar to the diameter of the eyepiece, the focal length of the eyepiece affects the magnification of the image produced by the objective. Changing the eyepiece's focal length changes magnification but does not change the resolving power, which is determined by the objective. This option is incorrect.
  4. Diameter of Objective lens: If the diameter ($D$) of the objective lens increases, the minimum resolvable angle ($\Delta \theta$) decreases ($\Delta \theta \propto 1/D$). A smaller minimum resolvable angle means the telescope can distinguish between objects that are closer together, thus increasing the resolving power. This option is correct.

Conclusion

To increase the resolving power of a telescope, the most effective method among the options is to increase the diameter of the objective lens. A larger objective lens collects more light and reduces the effects of diffraction, allowing for finer detail and better separation of close objects.

Revision Table: Factors Influencing Telescope Performance

Factor Affects Resolving Power? Relationship with Resolving Power Affects Magnification? Relationship with Magnification (Simple Refracting Telescope)
Diameter of Objective Lens ($D$) Yes Resolving Power $\propto D$ No N/A
Wavelength of Light ($\lambda$) Yes Resolving Power $\propto 1/\lambda$ No N/A
Focal Length of Objective Lens ($f_o$) No N/A Yes Magnification $\propto f_o$
Focal Length of Eyepiece ($f_e$) No N/A Yes Magnification $\propto 1/f_e$
Diameter of Eyepiece No N/A No (affects field of view) N/A

Additional Information on Telescope Performance

Resolving Power vs. Magnification

It's important to distinguish between resolving power and magnification. Magnification makes objects appear larger, but if the telescope doesn't have enough resolving power, the magnified image will simply be a larger blur. High resolving power is essential for seeing fine details, regardless of how much the image is magnified.

The Role of Objective Diameter (Aperture)

The diameter of the objective lens (aperture) is the single most critical factor determining both the resolving power and the light-gathering ability of a telescope. A larger aperture collects more light, allowing you to see fainter objects, and provides higher resolving power, allowing you to see finer details.

Practical Limitations

While theoretically, decreasing wavelength or increasing diameter increases resolving power, there are practical limits. Using shorter wavelengths like UV or X-rays requires different detector technology and telescopes must be placed outside the Earth's atmosphere. Increasing the diameter of the objective lens makes the telescope significantly larger, heavier, and more expensive to build and maintain. Atmospheric turbulence ("seeing") also limits the achievable resolving power from ground-based telescopes, often below the theoretical limit set by the objective diameter.

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Important Questions from Ray Optics and Optical Instruments

  1. A Convex mirror produces the magnification 1/3 and 1/4 when the object is placed at the points P and Q in front of the mirror.

  2. Which of the following statements are correct?

    • A. The saturation current is constant with collector plate potential for different frequencies of incident radiation.
    • B. The saturation current is different with collector plate potential for different frequencies of incident radiation.
    • C. The saturation current is different with collector plate potential for different intensity of incident radiation.
    • D. The saturation current is constant with collector plate potential for different intensity of incident radiation.
    • E. Below threshold frequency, no photoelectrons are emitted.

    Choose the correct answer from the options given below:

  3. For insulators and semiconductors, the resistance decreases with an increase in temperature because:

  4. A ray of light passes through four transparent media with refractive index μ1, μ2, μ3, and μ4 as shown in the figure. The surfaces of all media are parallel. If BC and DE are parallel, we must have:

  5. Light of uniform intensity shines perpendicularly on a totally absorbing surface, fully illuminating the surface. If the area of the surface is decreased, what is the effect on radiation pressure?

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