CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

With a concave mirror, an object is placed at a distance $${ X }_{ 1 }$$ from the principle focus, on the principle axis. The image is formed at a distance $${ X }_{ 2 }$$ from the principle focus. The focus length of the mirror is


A
x1x2
loader
B
x1+x22
loader
C
x1x2
loader
D
x1x2x1+x2
loader

Solution

The correct option is A $$\sqrt { { x }_{ 1 }{ x }_{ 2 } } $$
$$\begin{array}{l}\text { Here } u=\left(f+x_{1}\right) \\ v=-\left(f+x_{2}\right) \\\text { from mirror formula } \\\qquad \begin{aligned} -\frac{1}{f} &=\frac{-1}{\left(f\left(+x_{2}\right)\right.}+\frac{-1}{\left(f+x_{1}\right)} \\\frac{1}{f} &=\frac{f+x_{1}+f+x_{2}}{\left(f+x_{2}\right)\left(f+x_{1}\right)}\end{aligned}\end{array}$$
$$\begin{array}{c}\Rightarrow 2 f^{2}+\left[\left(x_{1}+x_{2}\right)=f^{2}+f\mu_{1}+f x_{2}+x_{1} x_{2}\right. \\ f^{2}=x_{1} x_{2} \\ f=\sqrt{x_{1} x_{2}}\end{array}$$

2007846_1358329_ans_c2cb972ac2c74ff2a5d22d41ef61e132.PNG

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image