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Answer the following questions : (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provided angular magnification ? (b) In viewing through a magnifying glass, one usually positions one's eye very close to the lens. Does angular magnification change if the eye is moved back ? ( c) Magnifying power of a simple a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ? (d) Why must both the objective and eye piece of a compound microscope have a short focal lengths ? (e) when viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best veiwing, why ? How much should be that short distance betweenthe eye and eye piece ?

Answer» <html><body><p></p>Solution :(a) It is true that angular size of imahe is equal to angular size of the object. <br/> By using magnifying glass, we keep the object far more closer to the eye than at `25 cm`, its normal position without use of glass. The closer object has <a href="https://interviewquestions.tuteehub.com/tag/larger-1067345" style="font-weight:bold;" target="_blank" title="Click to know more about LARGER">LARGER</a> angular size than the same object at `25 cm`. It is in this sense that angular magnification is achieved. <br/> (b) Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The <a href="https://interviewquestions.tuteehub.com/tag/effect-966056" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECT">EFFECT</a> is negligible when image is at much larger distance. <br/> ( c) Theoretically, it is true. However, when we decrease focal length, aberrations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths. <br/> (d) Angular magnification of eye piece is `(1 + (d)/(f_e))`. This increases as `f_e` decreases. <br/> Further, magnification of objective lens is `(v)/(u)`. As object <a href="https://interviewquestions.tuteehub.com/tag/lies-1073086" style="font-weight:bold;" target="_blank" title="Click to know more about LIES">LIES</a> close to focus of objective lens `u ~~ f_0`. To increase this magnification `(v//f_0), f_0` should be smaller. <br/> (e) The image of objective lens in eye piece is called 'eye ring'. All the rays from the object <a href="https://interviewquestions.tuteehub.com/tag/refracted-7709131" style="font-weight:bold;" target="_blank" title="Click to know more about REFRACTED">REFRACTED</a> by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only. When eye is too close to the eye piece, <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a> of view reduces and eyes fo not collect much of the light. The precise location of the eye would depend upon the separation between the obeject and eye piece, and also on focal length of the eye piece.</body></html>


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