Question 22

[anuradharo8690]

Hi,

I am getting a bit confused about where the "v" for lens 1 comes from.

"v = 1 - 2 = -1 cm (since lenses 1 and 2 are 1 cm apart)"

Could you please clarify this?

Thanks.

[admin]

Whenever you have 2 lenses, you must consider the following: the second lens magnifies the image produced by the first lens. So when looking at lens 1 in the system in the passage, it will create an image 1 cm away at lens 2 (though this was calculated in the Explanation, this information is actually given in the question).

[ladywait223593]

So the diverging lens 1 creates an image 1 cm in front of the scope (to the left)? It's negative because it's an image and on the same side as the object. And this same position relative to the second lens is 2 cm to the left of its optic center and since it represents the object to lens 2 it's considered positive. If the gap between lens 1 and 2 had been 4 cm then the image location relative to lens 1 would have to be (4-2)=+2cm and therefore 2 cm to the right of lens 1. But this would have required lens 1 to be converging.

I think I get it. There's no way I would have figured this out in less than a minute though. The main thing is to tell yourself lens 2's object needs to be 2 cm in front of its center and since lens 1 is 1 cm in front of lens 2 then we can logically see how this puts that object only 1 cm in front of lens 1.

[dcherian8173]

Why was it necessary to calculate anything for lens 2? Since lens 1 is diverging, we know that all images it creates will be negative. And as you said, the distance between lens 1 and lens 2 (1cm) is given in the passage.

-1/2 = 1/d + (-1/1); d = 2cm

[prestonker2934]

You can't magnify a virtual image, therefore lens 1 cannot be a diverging lens. Its a convex lens