There are two lenses. Lens 1 is a converging lens. (local length + 18 cm). Len 2

Question

There are two lenses. Lens 1 is a converging lens. (local length + 18 cm). Len 2 is a converging lens (focal length + 21 cm) placed 3 cm to the right of lens 1. An object is placed 6 cm to the left of lens 3. Determine if the final image is real or virtual. Also determine if the final image is inverted or upright as compared to the initial object. a) Real Inverted b) Real Upright c) Virtual Inverted d) Virtual Upright There are two lenses. Lens 1 is a converging lens (focal length +15 cm). Lens 2 is a converging lens (focal length +10 cm) placed 10 cm to the right of lens 1. An object is placed 30 cm to the left of lens 1. Determine if the final image is real or virtual. Also determine if the final image is inverted or upright as compared to the initial object. a) Real Inverted b) Real Upright c) Virtual Inverted d) Virtual Upright

Explanation / Answer

The method to attempt all 3 questions is same.

Since question 8 is the clearest, I'll start with it.

We'll start with lens 1 first.

u1= - 30

f1= +15

Using the lens' formula,

1/f=1/v-1/u

So, v= uf/ (u+f)

v1= 15 x -30 / -15 = 30 cm to the right of lens 1.

For the second lens, the object will be a virtual object, placed 30-10 = 20 cm to the right of lens 2.

So, u2= +20

f2= +10

v2= 20 x 10 / 30 = 20/3 cm to the right of lens 2.

Now to determine other factors, we'll see it's magnification which is v/ u for a lens.

m1 (due to lens 1) = 30 / -30 = -1

m2 ( due to lens 2)= 20 / (20 x 3) = 1/3

m1 x m2 = -1/3.

So, the final iamge will be real (due to negative sign of magnification)

and inverted, because the ratio h2/h1 = -1/3 and if h1 is positive then h2, the height of image, is negative which means inverted image is formed.

Also, two convergent lenses placed together will always give you a real, inverted image, as long as the setup is in air.

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