# Why does the definition of angular magnification relative to the substended angle of least distence that the eye can see

The angular magnification defined as,

angle subtended by image of object /angle subtended by object when 25 cm from the naked eye

Consider this.

An object is placing 50cm in front of a convex lens which has a focus length of 100cm. So the image will create 100cm in front of the lens.

If we consider the angular magnification,

angle subtended by image of object /angle subtended by object when 25 cm from the naked eye

=(h/100)/(H/25) [h=image height H=object height]

My question is this. According to the above definition this should be correct. But, according to my feeling, I feel that the meaning of the angular magnification of the lens is,

angle subtended by image of object /angle subtended by object at its own true placement

(That because according to the word 'angular magnification' I feel it is relative to its normal location angle, that is not 25cm, it is regarding to the place which is keeping the object)

That means according to above example, =(h/100)/(H/50)

What is the wrong in my thinking. Why do the definition limited the object to the least distance of eye can clearly see? Why isn't it depend on the place which the object keeps...... I am confuse with them.

As an object come closer to the unaided eye the image of the object focussed on the retina becomes larger, so the object is “seen” as being larger.

However there is a minimum distance between object and eye for which the eye can produce a focussed image on the retina.
The object is then said to be at the near point of the eye and the distance between the object and the eye is called the least distance of distinct vision.
When the object is at the near point your perceive the object to be of maximum size.
The least distance of distinct vision varies from person to person so a distance of $25\, \rm cm$ has been chosen as a reference distance for the “average” eye.

The least distance of distinct vision is then used as a reference to evaluate how much larger an image an optical instrument can produce than the best the unaided eye can do.

• Thanks for the answer, but I have a small doubt that, when the object keeps with a lens, what is the wrong, if we can define it as, how much larger an image an optical instrument can produce than the object distance with the lens that the unaided eye can see. Then we can get a full relation with the lens properties. But why doesn't it defined as that generally? – Osal Selaka Jul 9 '18 at 17:35
• @OsalSelaka There is nothing wrong with your definition other than the fact that very few (nobody else?) uses that definition. – Farcher Jul 9 '18 at 21:05
• Am I correct if I say like this, it is difficult to compare the sizes of the images, if I use my definition. So, using a standard(25cm angle) gives a opportunity for compare the sizes of the images. Is it the reason for using a standard? – Osal Selaka Jul 10 '18 at 4:38
• @OsalSelaka All scientists agree to make the comparison with the best the unaided eye can do when the object is 25 cm from the unaided eye and as far as I know only you have suggested a different distance, – Farcher Jul 10 '18 at 5:53