A ray of light traveling with speed c leaves point 1 shown in the figure and is

Question

A ray of light traveling with speed c leaves point 1 shown in the figure and is reflected to point 2. The ray strikes the reflecting surface a horizontal distance from point 1.

(a)

(b)

(c)

(d)

A ray of light traveling with speed c leaves point 1 shown in the figure and is reflected to point 2. The ray strikes the reflecting surface a horizontal distance from point 1. (a) t = frac{{sqrt {y_1 ^2 + x^2 } - sqrt {y_2 ^2 - left( {l - x} right)^2 } }}{c} (b) t = frac{{sqrt {y_1 ^2 - x^2 } + sqrt {y_2 ^2 - left( {l + x} right)^2 } }}{c} (c) t = frac{{sqrt {y_1 ^2 - x^2 } - sqrt {y_2 ^2 + left( {l + x} right)^2 } }}{c} (d) t = frac{{sqrt {y_1 ^2 + x^2 } + sqrt {y_2 ^2 + left( {l - x} right)^2 } }}{c}

Explanation / Answer

For this problem you need to use two formulas: time = distance / velocity pythagorean theorem sqrt( a^2 + b^2 ) = c for a right triangle where c is the hypotenuse The distance the light beam travels is the hypotenuses of the two right triangles so we need to find the length of the two hypotenuses and then add them together, for the triangle on the left we get: Hypotenuse = sqrt ( x^2 + y1^2) For the second triangle on the right we get: Hypotenuse = sqrt( (l-x)^2 + y2^2) we get l-x because the bottom leg of the right triangle is the total length l minus the bottom leg of the left triangle. So adding these two together we get that the total distance traveled is equal to: d = sqrt( y1^2 + x^2) + sqrt( (l-x)^2 + y2^2 ) Now, using the formula time = distance / velocity we get t = sqrt( y1^2 + x^2) + sqrt( (l-x)^2 + y2^2 ) / c where c is the speed of light, or answer D

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