Question 15 Marks
i. State laws of reflection of light.
ii. An object of height 5.0 cm is placed at 15 cm in front of a concave mirror of focal length 10 cm . At what distance from the mirror should a screen be placed, so that a focussed image is obtained on it? Find the height of the image.
ii. An object of height 5.0 cm is placed at 15 cm in front of a concave mirror of focal length 10 cm . At what distance from the mirror should a screen be placed, so that a focussed image is obtained on it? Find the height of the image.
Answer
View full question & answer→i. The angle of incidence is equal to the angle of reflection.The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane.
ii. u = -15 cm, f = -10 cm (concave mirror)
$\mathrm{h}=5.0 \mathrm{~cm}$
Mirror formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{v}=\frac{-1}{10 \mathrm{~cm}}+\frac{1}{15 \mathrm{~cm}}=\frac{-1}{30 \mathrm{~cm}}$
or $\mathrm{v}=-30 \mathrm{~cm}$. The screen must be placed at a distance of 30 cm from the mirror in front of it
$(m)=\frac{h^{\prime}}{h}=-\frac{v}{u}$
$h=\frac{-\dot{v}}{u} \times h=-\frac{-30 \mathrm{~cm}}{-15 \mathrm{~cm}} \times 5 \mathrm{~cm}=-10 \mathrm{~cm}$
ii. u = -15 cm, f = -10 cm (concave mirror)
$\mathrm{h}=5.0 \mathrm{~cm}$
Mirror formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{v}=\frac{-1}{10 \mathrm{~cm}}+\frac{1}{15 \mathrm{~cm}}=\frac{-1}{30 \mathrm{~cm}}$
or $\mathrm{v}=-30 \mathrm{~cm}$. The screen must be placed at a distance of 30 cm from the mirror in front of it
$(m)=\frac{h^{\prime}}{h}=-\frac{v}{u}$
$h=\frac{-\dot{v}}{u} \times h=-\frac{-30 \mathrm{~cm}}{-15 \mathrm{~cm}} \times 5 \mathrm{~cm}=-10 \mathrm{~cm}$
