A concave and convex lens of sa each having same focal length
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Then both have same power of magnitude..
but opposite sign
but opposite sign
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I have a great deal of sympathy with your position. As taught in schools the sign convention is somewhat vague and that gets confusing with complicated setups.
If you're dealing with anything complicated I recommend keeping strictly to the Cartesian sign convention. I've linked an article that seems to be a good summary of this, but Googling will find you lots more articles to look at. Basically:
the light travels from left to right
to the left of the lens is negative
to the right of the lens is positive
converging lenses have positive ff, diverging lenses have negative ff
This convention means uu is normally negative because the object is to the left of the lens, while vv is normally positive because a (real) image is to the right of the lens. The lens equation becomes:
1v−1u=1f.
If you're dealing with anything complicated I recommend keeping strictly to the Cartesian sign convention. I've linked an article that seems to be a good summary of this, but Googling will find you lots more articles to look at. Basically:
the light travels from left to right
to the left of the lens is negative
to the right of the lens is positive
converging lenses have positive ff, diverging lenses have negative ff
This convention means uu is normally negative because the object is to the left of the lens, while vv is normally positive because a (real) image is to the right of the lens. The lens equation becomes:
1v−1u=1f.
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