Focal lengths of two thin lenses kept in contant are `f_1` and `f_2`. Prove that their equivalent focal length f is given by `frac(1)(f)= frac(1)(f_1)+ frac(1)(f_2)`

Let's simplify : 2frac(1)(3) div 1frac(1)(6) div 1frac(1)(4)

What will be the equivalent focal length for two lenses of focal length f_1 and f_2 in contact ?

The focal length of a concave mirror is f. A point object is placed beyond the forcal length at a distance xf from the focus. Prove that the image will be formed beyond the focal length at a distanfe (f)/(x) from the focus and the magnification of the image will be -(1)/(x) .

At the pointx = 0, for the function f(x) = frac(1)(2-|x| the following are true ?

Simplify - (xii) 1frac(1)(2) [ 3frac(1)(2) + 2frac(1)(3) { 1frac(1)(4) + (2+ 3frac(2)(3) )}]

Two identical equiconvex lenses, each of focal length f, are placed side by side in contact with each other with a layer of water in between them as shown in the figure. If refractive index of the material of the lenses is greater than that of water, how the combined focal length F is related to f?

Let f be continuous on the interval [0,1] to R such that f(0)=f(1)dot Prove that there exists a point c in [0,1/2] such that f(c)=f(c+1/2)dot

If 'f' is the focal length of a lens then prove uv=f^2 . U and V represent object and image distances from focus respectively.

f(x) = sin^-1 frac(2x)(1+x^2) then