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):-

Options

A. Final image is formed in the plane of converging lens
B. If `x gt 1.5d` the rays will not be reflected by converging mirror
C. For `x gt d` final real images can be captured on a screen after on optical event (i.e refraction or reflection)at every optical system
D. Final image is real and inverted

Answer

A, C, D

Explanation

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
`i_(1)=(O_(1)f_(1))/(O_(1)-f_(1))=(-4.4)/(4-(-4))dm=-2dm` (dmrarrdecimeter)
Since the linear magnification is
`m_(1)=(H_(i1))/(H_(O1))=(i_(1))/(O_(1))=(1)/(2)`
the height of the first image is
`H_(i1)=(x)/(2)=H_(O2)`.
This virtual image is seen by the diverging lens as an object with an object distance of `O_(2)=f_(2)+|i_(2)|=6dm`, therefore the image formed by second lens has an image distance of
`i_(2)=(O_(2)f_(2))/(O_(2)-f_(2))=(6.4)/(6-4)dm=12dm`
The linear magnification of the second lens is
`m_(2)=(i_(2))/(O_(2))=(12)/(6)=2`

Hence the height of the second image is `H_(i2)=m_(2)H_(O2)=m_(2)H_(i1)=xc=H_(O3.)`
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
`i_(3)=(O_(3)f_(3))/(O_(3)-f_(3))=(-8.8)/(-8-8)=(-64)/(-16)=1dm`
This is a real image and since the linear magnification of the mirror is:
`m_(3)=(i_(3))/(O_(3))=(1)/(2)`
the height of the final image is
`H_(i3)=m_(3)H_(O3)=(x)/(2)`
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 rays will all pass under the concave mirror and there will be no image formation. Therefore the solution of the problem is :
`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:
`d lt x lt 1.5 d`.