Question

1) An object is placed 10cm in front of a concave mirror of
radius of curvature 16cm. The image will be formed
(a) + 30cm (b)-30cm (c) +40cm (d) - 40cm.
​

Answer 1

AnswEr:

Option (D) is correct

Step By Step Explanation -

Let us Consider that the radius of curvature of the mirror is - 16 cm.

And, Distance of the Image is - 10cm.

$:\implies\sf\; u = - 10 cm$

$:\implies\sf\; r = -16 cm$

$\rule{200}1$

$\dag\bold{\underline{\sf{\red{Now\; By\; Using\; Mirror\; Formula}}}}$

★ $\large{\underline{\boxed{\sf{\red{\dfrac{1}{v} \; + \; \dfrac{1}{u} \; = \; \dfrac{2}{r}}}}}}$

Now, Putting Values

$:\implies\sf\; \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{r}$

$:\implies\sf\; \dfrac{1}{v} + \dfrac{\; \; 1}{- 10} = \dfrac{\;\;2}{-16}$

$:\implies\sf\; \dfrac{1}{v} = \dfrac{\;\; 1}{10} - \dfrac{\;\;2}{16}$

$:\implies\sf\; \dfrac{16 \; - \; 20}{160}$

$:\implies\sf\; \dfrac{1}{v} = \dfrac{-4}{160}$

$:\implies\sf\; \dfrac{1}{v} = \dfrac{\;\;1}{- 40}$

$:\implies\large{\underline{\boxed{\sf{\pink{v \; = \; -40\; cm}}}}}$

$\rule{200}1$

The Image is placed at 30 cm in front of concave mirror & The formed image will be real, Inverted & magnified.