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The situation is shown below. Use Snell's Law \(\mathrm {1.5\; \sin\; \theta = 1.33\; \sin\; 90}\) \(\mathrm{\sin\; \theta = 0.887}\) \(\mathrm {\theta = 62.5^\circ}\)

(a) \(\mathrm {p = 20.0 \;cm, f = 12.0 \;cm}\) \(\mathrm {1/p + 1/q = 1/f}\) \(\mathrm {1/20 + 1/q = 1/12}\) \(\mathrm {1/q = 1/12 - 1/20 = (20 - 12)/(12\times20) = 8/240}\) \(\mathrm {q = 240/8 = 30.0\; cm}\) Now find the magnification: \(\mathrm {m = -q/p = -30/20 = -1.5}\) i.e., Image is magnified and inverted as in the figure. (b)  \(\mathrm {p = 20.0 \;cm, f = -12.0 \;cm}\) \(\mathrm {1/p + 1/q = 1/f}\) \(\mathrm {1/20 + 1/q = 1/-12}\) \(\mathrm {1/q = -1/12 - 1/20 = -32/240}\) \(\mathrm {q = -240/32 = -7.5 \;cm.}\) (Image is upstream and virtual) Now find the magnification: \(\mathrm {m = -q/p = -(-7.5)/20 = 0.375}\) i.e., Image is demagnified and erect as in the figure. (c) ​​​​​​​ \(\mathrm {p = 10.0 \;cm, f = 12.0 \;cm}\) \(\mathrm {1/p + 1/q = 1/f}\) \(\mathrm {1/10 + 1/q = 1/12}\) \(\mathrm {1/q = 1/12 - 1/10 = -2/120}\) \(\mathrm {q = -120/2 = -60 \;cm.}\) Image is upstream and virtual. (Object is inside the focal length and the lens is operating as a simple magnifying glass) Now find the magnification: \(\mathrm {m = -q/p = -(-60)/10 = 6}\) i.e., Image is magnified and erect.

The performance of retroreflectors is characterized within a geometrical coordinate system, usually with three angles for the incident and viewing geometries and a fourth orientation angle for prismatic designs like corner cubes, which are not rotationally isotropic in their performance. All the geometric variations are described in detail in ASTM E808-01, Standard Practice for Describing Retroreflection, along with expressions for converting from one geometric system to another.

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B is the correct situation from Q1 Which is the correct equation to use in this case relating: \(\mathrm {\theta_1, \; \theta_2, \; n_1, \;n_2?}\) \(\mathrm {n_1 \; \sin \; \theta_1 = n_2 \; \sin \; \theta_2}\)

\(\mathrm {1/p + 1/q = 1/f}\) \(\mathrm {1/-25 + 1/q = 1/-10}\) \(\mathrm {1/q = -1/10 + 1/25 = (10 - 25)/250 = -15/250}\) \(\mathrm {q = -250/15 = -16.7 \;cm}\)

Baroptics

p - POSITIVE The object is measured upstream from the vertex. q - POSITIVE Image distances are measured positive downstream from the vertex. R - NEGATIVE Radii are measured positive pointing upstream.

p - POSITIVE Object distances are measured positive upstream. q - NEGATIVE Image distances are measured positive downstream. f - POSITIVE Converging lenses have f positive.

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\(\mathrm {n_2 \sin \; \theta_3 = n_3 \sin\; \theta_4}\) \(\mathrm {1.5\; \sin\; 19.5 = 1.3\; \sin \; \theta_4}\) \(\mathrm {\sin\; \theta_4 = (1.5)(0.333)/(1.3) = 0.385}\) \(\mathrm{\theta_4 = 22.6^\circ}\)

\(\mathrm {1\; \sin\; 30 = 1.5\; \sin\; \theta_2}\) \(\mathrm {\theta_2 = 0.5/1.5 = 0.333}\) \(\mathrm {\theta_2 = 19.5^\circ.}\)

Retroreflectors are often specified by the coefficient of retroreflection, RA, for various observation angles and entrance angles.

Recall that a light ray bends towards the normal when it passes from a low index of refraction (\(\mathrm {n_1 = 1.0}\), in this case) medium to a higher index of refraction \(\mathrm {(n_2 = 1.5)}\) medium.  In the figure, only the rays in (b) and (d) bend towards the normal at the upper interface.

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Another useful angle for interpreting the performance of retroreflectors is the viewing angle, υ, the angle between the viewing direction and the normal to the retroreflector surface.

Waveoptics

Now find the magnification For 1st lens: \(\mathrm {p_1 = 15 \;cm, \;q_1 = 30 \;cm}\) For 2nd lens: \(\mathrm {p_2 = -25 \;cm, \;q_2 = -16.7 \;cm}\) \(\mathrm {m_1 = -q_1/p_1 = -30/15 = -2}\) \(\mathrm {m_2 = -q_2/p_2 = -(-16.7)/(-25) = -0.668}\) \(\mathrm{m = m_1m = (-2)(-0.668) = 1.34}\)

p - POSITIVE Object distances are measured positive upstream. q - POSITIVE Object distances are measured positive upstream. f - POSITIVE Converging lenses have f positive.

p - POSITIVE Object distances are measured positive upstream. q - POSITIVE Converging lenses have f positive. f - NEGATIVE Diverging lenses have f negative.

Values for RA of several hundred (cd/m2)/lux are not uncommon, corresponding to reflectance factors up to and over 1000.

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where E┴ is the illuminance on a plane normal to the direction of illumination, and I is the intensity of the illuminated retroreflector.

Two angles commonly used to specify the performance of retroreflectors are the entrance angle, ß, and the observation angle, α. The entrance angle is the angle between the illumination direction and the normal to the retroreflector surface. High-quality retroreflectors work over fairly wide entrance angles, up to 45-deg or more (up to 90 deg for pavement marking). The observation angle, the angle between the illumination direction and the viewing direction, is generally very small, often one degree or less.

Retroreflectors reflect incident light back toward the direction of the light source, operating over a wide range of angles of incidence. Typically they are constructed in one of two different forms, 90-deg corner cubes or high index-of-refraction transparent spheres with a reflective backing. Retroreflectors are used in transportation systems as unlighted night-time roadway and waterway markers, as well as in numerous optical systems, including lunar ranging. Some are made of relatively inexpensive plastic pieces or flexible plastic sheeting, and some are made of high-priced precision optics.

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At the lower interface, the light ray is passing from a higher index of refraction \(\mathrm {(n_2 = 1.5)}\) medium to a lower index of refraction \(\mathrm {(n_3 = 1.3)}\) material.  In this case, the light ray should bend away from the normal at the lower interface.  In (b), the light ray bends away from the normal, whereas the light ray in (d) bends towards the normal.

Therefore the final image is 16.7 cm to the left (upstream) of the 2nd lens or 16.7 - 5 = 11.7 cm to the left of the 1st lens.

p - POSITIVE The object is measured upstream from the vertex. q - NEGATIVE Image distances are measured positive downstream from the vertex. R - POSITIVE Radii are measured positive pointing upstream.

Note: If you wish further help on Snell's Law, you may find this bending light video entertaining and useful. The figures below correspond to questions 1-3.

A negative value for q means that the distance from the curved surface of the paperweight to the image is against the flow of light. That is, the image is 4 cm to the left of the curved surface of the paperweight.

p - POSITIVE The object is measured upstream from the vertex. q - POSITIVE Image distances are measured positive downstream from the vertex. R - POSITIVE Radii are measured positive pointing upstream.

(a) We must first solve for the 1st lens. \(\mathrm {1/p + 1/q = 1/f}\) \(\mathrm {1/15 + 1/q = 1/10}\) \(\mathrm {1/q = 1/10 - 1/15 = 5/150}\) \(\mathrm {q = 150/5 = 30 \;cm.}\) ​​​​​​​The image formed by the 1st lens is 30 cm downstream (to right) from the lens. Therefore it is 25 cm downstream from the 2nd lens. The image formed by the first lens becomes the object for the second lens. (b) NEGATIVE For the 2nd lens: \(\mathrm {p = -25 \;cm, f = -10 \;cm}\)