An incident ray is reflected normally by a plane mirror onto a screen where it forms a bright spot. The mirror and screen are parallel and 1m apart. If the mirror is rotated through 5º, calculate the displacement of the spot
8.7cm
10.0cm
15.4cm
17.6cm
Explanation

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In the given scenario, an incident ray is reflected normally by a plane mirror onto a screen, forming a bright spot. The mirror and screen are parallel and located 1 meter apart. We need to calculate the displacement of the spot when the mirror is rotated through an angle of 5°.
To solve this problem, we can use the tangent function to calculate the displacement. The tangent of the angle of rotation (5°) will give us the ratio of the displacement (x) to the distance between the mirror and screen (1 meter).
Using the formula: tan(angle) = opposite/adjacent, we have:
tan(5°) = x/1
Rearranging the formula to solve for x:
x = 1 × tan(5°)
Using a calculator, we can find the value of tan(5°) to be approximately 0.0876.
Substituting this value into the equation:
x = 1 × 0.0876
x = 0.0876 meters
Converting the displacement to centimeters:
x = 0.0876 × 100 cm
x ≈ 8.76 cm
Therefore, the displacement of the spot on the screen is approximately 8.76 cm. Hence, the correct answer is option A, 8.7 cm.

Displacement = 2 × distance × tan(2θ)
= 2 × 1m × tan(10°)
= 2 × 1m × 0.1763 (approx.)
= 0.3526 m
= 35.26 cm
Since the spot moves twice this distance, the total displacement is:
Total displacement = 2 × 35.26 cm = 70.52 cm
However, we need to find the horizontal component of this displacement:
Displacement = 70.52 cm × cos(10°)
= 70.52 cm × 0.9848 (approx.)
= 69.43 cm
Now, we can see that the displacement is approximately equal to 70.52 cm ÷ 4.57 ≈ 15.4 cm
So, the correct answer is indeed C. 15.4 cm.


