Question

A covex and concave mirror each of focal length f are placed at a distance 4 f apart as shown in the fig. at what point on the principle axis of the two. mirror should a point source of light be placed for the raise of converge at the same point after being reflected first from the convex mirror and then from the concave mirror

Answer 1

Answer:

Distance of object from convex mirror is

$x = f + \sqrt3 f$

Explanation:

let the distance from convex mirror is "x"

now the image formed at

$\frac{1}{v} + \frac{1}{-x} = \frac{1}{f}$

so we have

$v = \frac{fx}{f + x}$

now the object distance from concave mirror is given as

$u = 4f + \frac{fx}{f + x}$

now its image must form at

$v = 4f - x$

now by mirror formula

$\frac{1}{4f - x} + \frac{1}{4f + \frac{fx}{f + x}} = \frac{1}{f}$

$\frac{f + x}{4f^2 + 4fx + fx} = \frac{1}{f} - \frac{1}{4f - x}$

$\frac{f + x}{4f^2 + 5fx} = \frac{3f - x}{f(4f - x)}$

$(4f - x)(f + x) = (4f + 5x)(3f - x)$

$4f^2 + 4fx - fx - x^2 = 12 f^2 - 4fx + 15 fx - 5x^2$

$4x^2 - 8fx - 8f^2 = 0$

$x^2 - 2fx - 2f^2 = 0$

$x = \frac{2f \pm \sqrt{4f^2 + 8f^2}}{2}$

$x = f + \sqrt3 f$

#Learn

Topic : combination of mirror

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