Answer the following questions:
Answer the following questions:

(a) An object subtends an angle at the eye which is equal to the angle subtended at the eye by the virtual image that is produced by a magnifying glass. Does the magnifying glass provide angular magnification? Explain.

(b)A person’s eyes are very close to the lens when he is viewing through a magnifying glass. Does angular magnification change if the eye is moved back?

(c) The focal length of the lens is inversely proportional to the magnifying power of a simple microscope. Why don’t we achieve greater and greater magnifying power by using a convex lens of smaller and smaller focal length?

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

(e) Our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing, when viewing from a compound microscope. Explain why? How much should be that short distance between the eye and eyepiece?

Answer :

(a) The image and the object have the same angular size. The objects can be viewed with the help of a magnifying glass when they are put at the shortest distance of clear vision. The angular size of the object grows as it gets closer. Because it gives angular magnification, a magnifying glass is employed. It’s tough to bring an object closer to the eye without magnification.

(b) Yes, the magnification angle changes. The angular magnification reduces as the distance between the eye and the magnifying glass grows. This is due to the fact that the angle occupied by the eye is smaller than the angle occupied by the lens. The image distance has no bearing on angular magnification.

(c) It’s tough to make lenses with short focal lengths. When the focal length is short, spherical and chromatic aberrations are more likely to occur. As a result, a convex lens’ focal length is not reduced in the same way.

(d) A compound microscope produces an angular magnification given by –

\left [ \left ( \frac{25}{f_{e}} \right )+1 \right ]

where,

fe represents focal length of the eyepiece

It can be concluded that if fe is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens for a compound microscope is given by the expression –

\frac{1}{(\left | u_{o} \right |f_{o})}

Where, uo is the object distance for the objective lens

fo is the focal length of the objective

When uo > fo, the magnification is large. The object is kept closer to the objective lens when using a microscope. As a result, the object distance is extremely short. As u0 drops, f0 decreases as well. As a result, both fe and f0 are small in the current case.

(e) When we put our eyes too close to the eyepiece of a compound microscope, we won’t be able to capture much-refracted light. As a result, the field of view has been reduced significantly. As a result, the image’s clarity is compromised. The ideal position for seeing via a compound microscope is with the eye-ring attached to the eyepiece. The distance between the objective lens and the eyepiece determines the precise location of the eye.