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Problem 100

# When looking at the reflection of an object in the mirror, it seems as if the object is actually behind the mirror. The place where the object seems to be is called the virtual image. In the accompanying figure, light from the object at point $$O$$ reflects off the mirror $$\underline{M N}$$ at point $$B$$ and strikes the eye at point $$D$$. Given (by the laws of physics) that the angle of incidence equals the angle of reflection (\angleOBM $\cong \angle \mathrm{DBN}$ ) and that the total distance from the eyes to the virtual image, I, must equal the total distance traveled by the light from the eyes to the object, show that the virtual image is diametrically opposite the object with respect to the mirror.

Expert verified
In summary, we have proven that the virtual image I is diametrically opposite the object O with respect to the mirror MN using law of reflection and geometric properties. We proved that $$\triangle BOD\cong\triangle BOA$$ by angle-angle-angle (AAA) congruence, which allowed us to show that the total distance from the eyes to the virtual image (DI) is equal to the total distance traveled by the light from the eyes to the object (BI). This confirms that the virtual image and the object are equidistant from the mirror and located on opposite sides, as required.
See the step by step solution

## Step 1: Draw the Figure and Denote Points

Draw the given figure with an object at point O, a mirror MN, and an eye at point D. Light from the object O reflects off the mirror at point B and strikes the eye at point D. Let I be the virtual image.

## Step 2: Connect DI and Extend to Point A

Draw a straight line from D to I. Extend the imaginary line from point O to point A on the opposite side of the mirror such that O and A are equidistant from the mirror, and line segment OD is parallel to line segment MN.

## Step 3: Prove Triangles BOD and BOA are Congruent

Since angles OBM and DBN are congruent (by the law of reflection), and angles OBD and DBA are both equal to 90° (since the mirror is perpendicular to our line of sight from eye to object), angle DBN and angle OBA are alternate interior angles for parallel lines OD and MN, with respective transversal line OB. As a result, ∠DBN $$\cong$$ ∠OBA. Since ∠OBD $$\cong$$ ∠DBA and ∠BOD $$\cong$$ ∠BOA, we can say that $$\triangle BOD\cong\triangle BOA$$ by angle-angle-angle (AAA) congruence.

## Step 4: Prove DI = OA

Since $$\triangle BOD\cong\triangle BOA$$, we can now say that BD = BA and OD = OA. The total distance from the eye to the object is BD + OD. Let's substitute BA for BD and OA for OD. So, the total distance from the eye to the object is BA + OA = BI. The total distance from the eye to the virtual image I is DI. Since BA + OA = BI, we can say that DI = BI, which implies that the total distance from the eyes to the virtual image (DI) is equal to the total distance traveled by the light from the eyes to the object (BI).

## Step 5: Conclude that the Virtual Image is Diametrically Opposite the Object with respect to the Mirror

Since we have proved that DI = BI, we can conclude that the virtual image I is diametrically opposite the object O with respect to the mirror MN. This is because the point at which the virtual image appears (I) is the same distance away from the mirror as the object O, and the total distance from the eye to the object is equal to the total distance from the eye to the virtual image.

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