Solved 2024 Sample Question Paper CBSE Class 10 Science — Question 6

(i) Given,

Object distance (u) = -5 cm

Radius of curvature (r) = 40 cm

Image distance (v) = ?

Focal length (f) = $\dfrac{r}{2}$ = $\dfrac{40}{2}$ = -20 cm [f in case of a concave mirror is negative.]

Using formula,

$\dfrac{1}{\text{f}}$ = $\dfrac{1}{\text{v}}$ - $\dfrac{1}{\text{u}}$

We get,

-$\dfrac{1}{\text{20}}$ = $\dfrac{1}{\text{v}}$ - $\dfrac{1}{\text{5}}$

$\dfrac{1}{\text{v}}$ = -$\dfrac{1}{\text{20}}$ + $\dfrac{1}{\text{5}}$

$\dfrac{1}{\text{v}}$ = $\dfrac{-1+4}{\text{20}}$ = $\dfrac{3}{20}$

So, v = $\dfrac{20}{3}$ = 6.67 cm

Hence, image is obtained at 6.67 cm behind the mirror.

(ii) Size of object (h_o) = 10 cm

Size of the image (h_i)= ?

From formula,

m = $\dfrac{h_i}{h_o}$ = - $\dfrac{\text{v}}{\text{u}}$

Substituting we get,

$\dfrac{h_i}{10}$ = $-\dfrac{6.67}{-5}$

h_i = 10 x $\dfrac{6.67}{5}$ = 13.34 cm

Hence, size of image is 13.34 cm

(iii) Ray diagram showing the formation of the image is given below: