Kvant Physics Problem 60

Verified: yes
Verdicts: PASS + PASS
Solve time: 11m50s
Source on kvant.digital

Problem

F60. The image of a candle is visible in a plane mirror. What will happen to it if a plane-parallel plate is placed between the mirror and the candle?

Setup and Assumptions

A candle is placed in front of an ideal plane mirror. Between the candle and the mirror a plane-parallel transparent plate of thickness $t$ and refractive index $n$ is inserted. The surrounding medium is air with refractive index approximately equal to $1$.

The quantity to be determined is the position of the visible image of the candle after insertion of the plate.

The mirror is assumed perfectly plane. The plate faces are parallel and perpendicular to the optical axis joining the candle and the mirror. The plate is transparent and nonabsorbing. Paraxial rays are considered, so only geometrical optics is required. Multiple reflections inside the plate are neglected because they affect only brightness, not the principal image position.

Let the distance from the candle to the mirror before insertion be $a$.

Physical Principles

The solution uses the laws of geometrical optics.

For a plane mirror, the image is located behind the mirror at a distance equal to the distance of the object from the mirror.

For a plane-parallel plate of thickness $t$ and refractive index $n$, viewed from air along a direction close to the normal, the apparent thickness is

$$t_{\text{app}}=\frac{t}{n}.$$

Equivalently, an object seen through the plate appears shifted toward the observer by

$$\Delta=t-\frac{t}{n} =t\left(1-\frac1n\right).$$

Since the plate has parallel faces, rays emerging from it remain parallel to the incident rays; only the apparent position of the object is changed.

Derivation

Consider first the mirror alone. If the candle is at distance $a$ from the mirror, its image is formed at distance $a$ behind the mirror.

Now insert the plane-parallel plate between the candle and the mirror.

The mirror observes the candle through the plate. Because of refraction, the candle appears to the mirror closer than its actual position. The reduction of the apparent distance is

$$\Delta=t\left(1-\frac1n\right).$$

Hence the candle's apparent distance from the mirror becomes

$$a' = a-\Delta = a-t\left(1-\frac1n\right).$$

The mirror forms an image of this apparent object. Consequently the image produced by the mirror lies behind the mirror at distance

$$a' = a-t\left(1-\frac1n\right).$$

This image is then observed through the same plate on the return path. The virtual image behind the mirror is seen through the plate and again appears shifted toward the observer by the same amount $\Delta$.

Thus the final visible image is displaced toward the mirror by an additional amount $\Delta$.

The total shift of the observed image relative to its original position is

$$\Delta_{\text{tot}} = 2\Delta = 2t\left(1-\frac1n\right).$$

Originally the image was at distance $a$ behind the mirror. After insertion of the plate it is seen at distance

$$a_{\text{new}} = a-2t\left(1-\frac1n\right)$$

behind the mirror.

Result

The image remains virtual and upright, but it moves toward the mirror.

The displacement of the visible image is

$$\boxed{\Delta_{\text{image}} = 2t\left(1-\frac1n\right)}.$$

The new distance of the image behind the mirror is

$$\boxed{ a_{\text{new}} = a-2t\left(1-\frac1n\right) }.$$

No numerical values are given in the problem statement, so a numerical substitution cannot be performed. The answer is completely determined by the plate thickness $t$ and refractive index $n$.

Sanity Checks

The quantity $2t(1-1/n)$ has dimensions of length, which matches the required dimension for an image displacement.

If $n=1$, the plate is optically indistinguishable from air. Then

$$\Delta_{\text{image}}=0,$$

so the image remains in its original position.

If the thickness tends to zero,

$$t\to0,$$

the displacement also tends to zero.

For any real transparent plate, $n>1$, hence

$$1-\frac1n>0,$$

and the image moves toward the mirror, which agrees with the fact that objects viewed through a refracting slab appear closer.

The most common error occurs when accounting for the two passages through the plate. The apparent shift $t(1-1/n)$ occurs once when the mirror "sees" the candle and once again when the observer sees the mirror image. Omitting one of these contributions gives an answer smaller by a factor of two.