Question #84df3

2 Answers
Feb 7, 2017

No shift.

Explanation:

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

resonance.ac.inresonance.ac.in

Feb 11, 2017

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)
drawndrawn
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)

drawndrawn

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.