Search any question & find its solution
Question:
Answered & Verified by Expert
Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eve and eveniece?
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eve and eveniece?
Solution:
1034 Upvotes
Verified Answer
(a) The angular size of the image is equal to the angular size of the object. By using a magnifying glass, the object can be placed much closer than $25 \mathrm{~cm}$. The closer object has larger angular size than the same object at $25 \mathrm{~cm}$. Thus angular magnification is achieved.
(b) The angular magnification changes, if the eye moved back.
It decreases a little because the angle subtended at the eye is then slightly less than angle subtended at the lens.
(c) First, it is difficult to grind lens of very small focal length. Secondly, if we decrease the focal length of a lens, both spherical and chromatic aberrations become more pronounced.
(d) Angular magnification of eye piece is $\left(1+\frac{\mathrm{d}}{\mathrm{f}_{\mathrm{e}}}\right)$ which is greater if $\mathrm{f}_e$ is smaller. Magnification of the objective lens is given by $\frac{\mathrm{v}_0}{\mathrm{u}_0}$. The object is held close the focus of the objective lens $\therefore \mathrm{u}_0 \approx \mathrm{f}_0$. Therefore magnification is $\frac{v_0}{f_0}$. As the $u_0$ is small so is $\mathrm{f}_0$.
(e) The image of the objective in the eye-piece is known as eye-ring'. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. The location of eyering depend on the separation between the objective and the eye piece.
(b) The angular magnification changes, if the eye moved back.
It decreases a little because the angle subtended at the eye is then slightly less than angle subtended at the lens.
(c) First, it is difficult to grind lens of very small focal length. Secondly, if we decrease the focal length of a lens, both spherical and chromatic aberrations become more pronounced.
(d) Angular magnification of eye piece is $\left(1+\frac{\mathrm{d}}{\mathrm{f}_{\mathrm{e}}}\right)$ which is greater if $\mathrm{f}_e$ is smaller. Magnification of the objective lens is given by $\frac{\mathrm{v}_0}{\mathrm{u}_0}$. The object is held close the focus of the objective lens $\therefore \mathrm{u}_0 \approx \mathrm{f}_0$. Therefore magnification is $\frac{v_0}{f_0}$. As the $u_0$ is small so is $\mathrm{f}_0$.
(e) The image of the objective in the eye-piece is known as eye-ring'. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. The location of eyering depend on the separation between the objective and the eye piece.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.