Question #84df3

Let three lenses we named as #L_1,L_2 and L_3# from left to right. Let Principal axis of all three be same, as shown in the figure below.

(1) Let the point source be placed at given distance of #10cm# to the left of #L_1#.

It is placed at the left Focus of this lens. All rays of light from the object will become parallel to the principal axis and strike lens #L_2#.

Lens #L_2# will form the image at its focus located to its right, i.e., #10cm# to right.

Since the distance between the lenses #L_1 and L_2# is #30cm#, the image formed by #L_2# is located at #20cm# to left of #L_3#.

This location of object for #L_3# is its left Center of curvature. Object at left #C# of lens #L_3# will form image at the other #C#, on the right.

Distance between Object and final image#=10+30+30+20=90cm#

(2) When the device is moved away from the source by another #10 cm#. Object is now placed at a distance of #20cm# to the left of #L_1#. This is its left Center of curvature. The image will be formed at the right Center of curvature, #20cm# to right of #L_1#.

Since the distance between the lenses #L_1 and L_2# is #30cm#,
this implies that It is located at the left Focus of lens #L_2#. All rays of light will become parallel to the principal axis and strike lens #L_3#.

Lens #L_3# will form the image at its focus located to its right, i.e., #10cm# to its right.

Distance between Object and final image#=20+30+30+10=90cm#

Given that the 3 convex lenses each of focal length #f=10cm#
are placed inside a hollow tube at equal spacing of #30 cm# each keeping their principal axis aligned as shown in the following figures.

We are to find out the distance between object and final image in two given situations.

Situation -1(Figure - 1)

In the 1st situation the point object is placed at a distance #10cm# from the lens i.e.on the focus of the objective lens. So the divergent beam of light after refraction through 1st lens will be transformed into parallel beam which after refraction through the 2nd lens will converge on focus of the 2nd lens and 1st real image will be formed there. This will be at #20cm=2f# apart from the third lens. This image will act as object point of third lens and the diverging beam of light emerging from it will meet after refraction through 3rd lens at a distance #2f=20cm# forming finally a 2nd real image

So it is obvious from figure that the distance between object and final image in the 1st situations is #(10+30+30+20)cm=90cm#

Situation -2 (Figure-2)

In the 2nd situation the point object is placed at a distance #20cm# from the lens i.e. at a point #2f# from the objective lens. So the divergent beam of light after refraction through 1st lens will converge at a point #2f=20cm# apart from 1st lens forming 1st real image there. This will be at #10cm=f# apart from the third lens. This 1st image will act as a point object of 2nd lens and the divergent beam of light emerging from it after refraction through 2nd lens will be transformed into parallel beam which after refraction through the 3rd lens will converge on focus of the 3rd lens and 2nd real image will be finally formed there.

So it is obvious from figure that the distance between object and final image in the 1st situations is #(20+30+30+10)cm=90cm#

Please note that the principle of reversibility of light also provides the same result as if in the second situation the object is placed in the final image position of the 1st situation and the image is formed in object position of the 1st situation. So their positions are interchangeable.