Chapter 34: Q7Q (page 1037)

The table details six variations of the basic arrangement of two thin lenses represented in Fig. 34-29. (The points labeled $F_{1}$ and $F_{2}$ are the focal points of lenses 1 and 2.) An object is distance $p_{1}$ to the left of lens 1, as in Fig. 34-18. (a) For which variations can we tell, without calculation, whether the final image (that due to lens 2) is to the left or right of lens 2 and whether it has the same orientation as the object? (b) For those “easy” variations, give the image location as “left” or “right” and the orientation as “same” or “inverted.”

Short Answer

Expert verified

For the given case variations $1, 3, 4, 5 & 6$ we can say that, image is towards left or right of lens.
Variation for $1, 3 & 4$ are right, real and inverted. Variation for $5, 6$ are Left, virtual, same.

Step by step solution

The given data

The two lens-systems with the object and focal distance relation for every variation are given.

Understanding the concept of lens

When a ray of light is passed through lens 1, the image formed depends on the position and the type of lens. If both sides of the lens curve are outward in form, the lens is called a converging lens, and here, the lens will bend light from distant objects inwards toward a single point, called the focal point. If both sides of the lens curve inward, it is called a diverging lens, and light from distant objects will bend outwards. As long as the object is outside of the focal point the image is real and inverted. When the object is inside the focal point the image becomes virtual and upright. Negative lenses diverge from the parallel incident light rays and form a virtual image by extending traces of the light rays passing through the lens to a focal point behind the lens.

a) Calculation of the variations that are either left or right of lens 2

The image distance due to lens 1 is $I_{1}$ .

The image distance due to lens 2 is $I_{2}$ .

The object distance from lens 1 is $p_{1}$ .

The object distance from lens 2 is $p_{2}$ .

Here, image formed by lensis the object for lens

Hence, the image distance from lens 1 is equal to the object distance of lens 2,

$i_{1} = p_{2}$

For converging lens, if

$p_{1} > f_{1}$ the image is real and inverted.
$p_{1} < f_{1}$ the image is virtual and upright.

For diverging lens, if

$p_{1} > f_{1}$ the image is virtual and located between the object and the lens.
$p_{1} < f_{1}$ the image is virtual and located between the object and the lens.

Variation 1:

lens1 : Converging

lens2: Converging.

$p_{1} < f_{1}$ gives a virtual and upright image to the left of lens

Hence, $p_{2} > f_{2}$ , so the converging lens gives a real and an inverted image to the right of the lens

Variation 2:

lens1 : Converging

lens2: Converging.

$p_{1} > f_{1}$ so, the converging lensgives a real and an inverted image to the right of lens

Now, we can’t say that whether this image is inside or outside of the focus of lens

Variation 3:

lens1 : Diverging

lens2: Converging.

$p_{1} < f_{1}$ so, the diverging lensgives a virtual and an upright image to the left of lens

Hence, $p_{2} > f_{2}$ , so the converging lensgives a real and an inverted image to the right of lens

Variation 4:

lens1 : Diverging

lens2: Converging.

$p_{1} > f_{1}$ , so diverging lensgives virtual and upright image to the left of lens

Hence, $p_{2} > f_{2}$ so the converging lensgives a real and an inverted image to the right of lens

Variation 5:

Lens1 : Diverging

Lens2: Diverging.

$p_{1} < f_{1},$ so the diverging lensgives a virtual and an upright image to the left of lens

Hence, $p_{2} > f_{2}$ so diverging lensgives a virtual and an upright image to the left of lens

Variation 6:

Lens 1: Diverging

Lens 2: Diverging.

$p_{1} > f_{1}$ , so the diverging lensgives a virtual and an upright image to the left of lens

Hence, $p_{2} > f_{2}$ , so the diverging lensgives a virtual and an upright image to the left of lens

From above data:

Hence, for variations $1, 3, 4, 5 & 6$ , we can say that the image is towards left or right of the lens.

b) Calculation of the positions and orientations of the variations

Now, from above calculations the positions and orientations of the variations, we can say that the image property of the variations 1, 3, 4, 5 and 6, while the image position for variation 2 is uncertain.

For variations 1, 3, and 4, we have that the image is real, inverted and upright as the object distance is greater than the focal length of the mirror for both the lens.

Similarly, for the variations 5,6, we have the image is towards the left of the lens with being virtual in nature and has same height that of the object with object distance less than the focal length for lens 1 while the object distance is greater than the focal length for lens 2.

Hence, variation $1, 3 & 4 :$ right, real and inverted and variation $5, 6$ : Left, virtual, same.