Recent content by nish95

Relative to the boat, the sail is oriented the same way in this and the previous question. One would think that the answer then, is that the boat sails at the same speed in both cases but that is not what the author says. I'm trying to understand why this is so. More details are in the original...

In the previous one, the boat is sailing vertically in the same direction as the wind.

Hi everyone, I was looking to develop my physical insight when I encountered this book by Lewis Caroll Epstein. For the crosswind problem, I couldn't understand the author's explanation; in particular, his concept of "artificial wind," and the force being larger in this case than the previous...

I find the convention that you are using really simple. Don't know why my book has to complicate things! Anyways, thanks for all the help. @TSny @hutchphd @ehild Btw, what does OP stand for ? Also, @TSny which textbooks do you guys use for high school physics?

It is to the right of the second lens. But rays have to travel backwards, that is towards the concave lens to form the final image. In that case my positive direction is right to left and distance is being measured from the optical center of the concave lens to the first image (left to right)...

I get what you are saying. But that doesn't change things. I mean the concave lens produces the final image nearer to itself than the first image made by the convex lens. So, magnitude of distance of final image should be less than 20 cm.

Yes, that's what I did but the book has taken it to be +ve.

Isn't the diagram correct? I mean rays coming from the virtual object will diverge because of the concave lens right?

I am using the convention that distances measured from optical centers are +ve in the direction of propagation of the ray and -ve if they both are opposite.

First image is an object for the concave lens so won't +ve direction change from right to left?! In that case, object distance will be -ve (from concave lens towards right side). Any ideas? Solution in the book takes first image's distance to be +ve. See attached ray diagram for clarification...

@TSny, thank you very much for clarifying!

See the solved example as shown in the image. I don't understand how can we write S(A)=2S(B) since integrating V(A)=2V(B) will give us an extra unknown constant and the work done by friction will depend on it. I found the relation 2S(B) + S(A) = const. (somebody confirm if this is right?) so...