Q.The magnifying power of an astronomical telescope for normal adjustment is -

What is Angular Magnification?

When you look at a tiny object — say a grain of salt — you hold it close to your eye to see it bigger. But there is a limit: bring it too close and it blurs. The closest distance at which your eye can focus comfortably is called the near point, conventionally taken as 25 cm for a normal eye. At that distance, the object subtends a certain angle at your eye. That angle determines how large it appears — not its physical size, but the fraction of your field of view it occupies.

Now imagine using a magnifying glass. The same grain of salt now looks much larger. Why? Because the lens lets you bring the object even closer than 25 cm while still seeing a clear, magnified image. That image is formed at a comfortable viewing distance, but the angle it subtends at your eye is far bigger than the angle the object would subtend at 25 cm without the lens.

Angular magnification is simply the ratio of these two angles:

Angular magnification $M = \frac{\theta_{\text{image}}}{\theta_{\text{object}}}$

where $\theta_{\text{image}}$ is the angle subtended by the image when viewed through the instrument, and $\theta_{\text{object}}$ is the angle subtended by the object when viewed with the naked eye at the near point (25 cm).

Why "Angular" and Not "Linear"?

A common confusion: a microscope or telescope does not give you a physically larger object — it gives you a larger apparent size. The image on your retina is bigger because the rays entering your eye are steeper. That steepness is measured by the angle. So magnification here is about angles, not actual lengths.

A Concrete Example

Take a simple magnifier (a convex lens) of focal length $f = 5$ cm. You place the object just inside the focal point so that a virtual, erect image forms at infinity (or at the near point). For the "image at infinity" case, the angle subtended by the image is $\theta_{\text{image}} \approx h / f$, where $h$ is the object height. The angle subtended by the object at the near point (25 cm) is $\theta_{\text{object}} \approx h / 25$.

Thus:

For $f = 5$ cm, $M = 5$. The image appears 5 times larger than the object seen at 25 cm.

The Big Picture

Angular magnification is the language of all optical instruments:

• Simple magnifier: $M \approx 25/f$ (image at infinity) …