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An astronomical telescope consists of two converging lenses. One of them of large aperture and large focal length is called objective lens and the other one, of smaller focal length and smaller - Physics

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प्रश्न

An astronomical telescope consists of two converging lenses. One of them of large aperture and large focal length is called objective lens and the other one, of smaller focal length and smaller aperture is called the eyepiece. It is used to see distant objects which are not seen clearly with naked eyes. The image formed by the objective lens acts as an object for the eyepiece and the final image produced by the eyepiece is magnified.
  1. The images formed by the objective lens and the eyepiece are respectively ______.    (1)
    1. virtual, real
    2. real, virtual
    3. virtual, virtual
    4. real, real
  2. The magnification produced by the telescope does not depend upon the ______.    (1)
    1. colour of light.
    2. focal length of objective lens.
    3. focal length of eyepiece.
    4. apertures of objective lens and eyepiece.
  3. Which of the following statements is not correct for this telescope?    (1)
    1. The focal length of objective lens (fo) is larger than the focal length of eyepiece (fe).
    2. Its magnifying power can be increased by increasing the focal length of objective lens (fo).
    3. The distance between two lenses is more than (fo + fe).
    4. The magnifying power can be decreased by increasing the focal length of eyepiece.

    1. An astronomical telescope has objective lens and eyepiece of focal lengths 80 cm and 4 cm respectively. To view the image in normal adjustment, the lenses must be separated by a distance of ______.    (1)
      1. 84 cm
      2. 76 cm
      3. 20 cm
      4. 320 cm
        OR
    2. Consider the telescope described in question (iv)(a). Its magnifying power in normal adjustment will be:    (1)
      1. 320
      2. 84
      3. 76
      4. 20
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उत्तर

i. The images formed by the objective lens and the eyepiece are respectively real, virtual.

Explanation:

The objective lens is the lens through which Light from an object at infinity enters the large objective lens. It converges the light to form a diminished, inverted, and real image at its second focal point.

The eyepiece lens intermediate real image is positioned within the focal length of the eyepiece. The eyepiece then forms a highly magnified, inverted (with respect to the original object), and virtual image.

ii. The magnification produced by the telescope does not depend upon the apertures of objective lens and eyepiece.

Explanation:

The focal lengths of the objective (fo) and eyepiece (fe) are clearly related to magnification. Because glass’s refractive index (and hence the focus length) varies with wavelength, it also depends on the color of light (Chromatic Aberration). The aperture (diameter) of the lenses influences both the light-gathering capacity and the resolving power (the sharpness or brightness of the image); however, it is not included in the mathematical equation for magnification. Thus, magnification is independent of the apertures of the lenses.

iii. The distance between two lenses is more than (fo + fe).

Explanation:

When normal adjustment is used (final picture at infinity), the distance is fo + fe. The eyepiece is moved closer to the objective, making the separation L = fo + ue, where ue is less than fe. This is done so that the final image is made at the shortest distance of distinct vision (D). In this case, L < fo + fe. In normal use, the distance between them is never more than fo + fe.

iv.

a. An astronomical telescope has objective lens and eyepiece of focal lengths 80 cm and 4 cm respectively. To view the image in normal adjustment, the lenses must be separated by a distance of 84 cm.

Explanation:

Given: Focal length of objective fo = 80 cm.

Focal length of eyepiece fe = 4 cm

In normal adjustment, the lenses are separated by the sum of their focal lengths:

L = fo + fe

= 80 cm + 4 cm

= 84 cm

∴ The separation distance is 84 cm.

OR

b. 20

Explanation:

From the previous part (iv)(a), we have:

Focal length of objective fo = 80 cm.

Focal length of eyepiece fe = 4 cm.

Calculating magnifying power:

M = `80/4`

= 20

This means the distant object will appear 20 times larger in angular size when viewed through the telescope.

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2025-2026 (March) 55/4/1
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