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InterviewSolution
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The following objects are placed one after each other in given order onto a central axis with a separation of 40 cm each. A point source of light O. a diverging lens of focal length 40 cm. a converging lens of focal length 40 cm and converging mirror of focal length 80 cm. The aperture diameter of lenses and mirror is d=20 cm. If a point source of light is placed at a perpendicular distance of x from central axis then. (You have to consider single optical event at any optical element) Mark the CORRECT statement(s):- |
| Answer» <html><body><p>Final image is formed in the plane of converging lens<br/>If `x gt 1.5d` the rays will not be reflected by converging mirror<br/>For `x gt d` final real images can be <a href="https://interviewquestions.tuteehub.com/tag/captured-7293063" style="font-weight:bold;" target="_blank" title="Click to know more about CAPTURED">CAPTURED</a> on a screen after on optical <a href="https://interviewquestions.tuteehub.com/tag/event-976579" style="font-weight:bold;" target="_blank" title="Click to know more about EVENT">EVENT</a> (i.e refraction or reflection)at every optical system<br/>Final image is real and inverted </p>Solution :Let us follow the way the images are formed by the lenses and the mirror. The virtual image formed by the diverging lens has an image distance of<br/> `i_(1)=(O_(1)f_(1))/(O_(1)-f_(1))=(-4.4)/(4-(-4))dm=-2dm`(dmrarrdecimeter) <br/> Since the linear magnification is <br/> `m_(1)=(H_(i1))/(H_(O1))=(i_(1))/(O_(1))=(1)/(2)`<br/> the height of the first image is <br/> `H_(i1)=(x)/(2)=H_(O2)`. <br/> This virtual image is seen bythe diverging lens as an object with an object distance of `O_(2)=f_(2)+|i_(2)|=6dm`, <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> the image formed by second lens has an image distance of<br/> `i_(2)=(O_(2)f_(2))/(O_(2)-f_(2))=(6.4)/(6-4)dm=12dm`<br/> The linear magnification of the second lens is <br/> `m_(2)=(i_(2))/(O_(2))=(12)/(6)=2`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ALN_XII_P4_E01_164_S01.png" width="80%"/> <br/> Hence the height of the second image is `H_(i2)=m_(2)H_(O2)=m_(2)H_(i1)=xc=H_(O3.)`<br/> The second image is seen by the concave mirror as a virtual object bering at a distance of `O_(3)=f_(3)-i_(2)=-8dm`, therefore the image is formed at <br/> `i_(3)=(O_(3)f_(3))/(O_(3)-f_(3))=(-8.8)/(-8-8)=(-64)/(-16)=1dm`<br/> This is a real image and since the linear magnification of the mirror is: <br/> `m_(3)=(i_(3))/(O_(3))=(1)/(2)`<br/> the height of the final image is <br/> `H_(i3)=m_(3)H_(O3)=(x)/(2)` <br/> Note that final image is formed in the plane of the converging lens, therefore the image can only be captured on a screen if its height is greater than the radius of the lens. This happens when x is greater than d, therefore the list condition for perpendicular is d lt x. If x is increased further than`1.5 d`,the rayswillall pass under the concave mirror and there will be no image formation. Therefore the solution of the <a href="https://interviewquestions.tuteehub.com/tag/problem-25530" style="font-weight:bold;" target="_blank" title="Click to know more about PROBLEM">PROBLEM</a> is : <br/> `d lt x lt 1.5 d`.creased further than `1.5 d`, the rays will all pass under the concave mirror and there will be no images formation. Therefore the solution of the problem is: <br/> `d lt x lt 1.5 d`.</body></html> | |