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Q3.

Expert-verifiedFound in: Page 583

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**A person with eyes at A, 150 cm above the floor, faces a mirror 1 m away. The mirror extends 30 cm above eye level. How high can the person see on a wall 2 m behind point A?**

The person can see **2.7** m high on a wall **2** **meter behind him**.

The eye of the person is at point *A*, 150 cm above the floor.

The wall is 2 m behind the person.

The mirror extends 30 cm above eye level.

The person faces mirror at a distance of 1 m.

Angle made by the ray connecting mirror and eye with horizontal

$\begin{array}{l}\mathrm{tan}\theta =\frac{\frac{30}{100}}{1}\\ \text{}=16.7\xb0\end{array}$

Thus angle made by the ray connecting mirror and eye with the normal of the mirror is also $16.7\xb0$(alternate interior angles between parallel lines are equal).

According to the law reflection the reflected ray will also make same angle with the normal of the mirror.

Therefore, angle of reflection is $16.7\xb0$.

Consider the height of wall above mirror level which the person can see be *x* m.

$\begin{array}{c}\mathrm{tan}16.7\xb0=\frac{x}{3}\\ 0.3=\frac{x}{3}\\ x=.9\text{}\left[\text{multiply each side by 3}\right]\end{array}$

The height of wall from floor level which the person can see is

$\frac{150}{100}+\frac{30}{100}+0.9=2.7$

Therefore the person can see 2.7 m high on a wall 2 meter behind him.

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