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

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.

Waveoptics

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

Opticsmeaning political

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.

Baroptics

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.

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.

optics期刊

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

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

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.

(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: \(\mathrm {p = -25 \;cm, f = -10 \;cm}\)

Optical lens

History ofoptics

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

Machine Shop & Requisition Form Poster Boards NMR Centre The Observatory is not available for public visits at this time.

Optica

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

Department Chair: Dr. Stefan Kycia Administrative Officer: Rachel Baker IT Help: cepsit@uoguelph.ca Other Inquiries: physinfo@uoguelph.ca

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.

Optics

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

Campus Directory Gryph Mail Physics Intranet Employee Portal Illness or Injury Incident Report Undergraduate Calendar Graduate Calendar Academic Integrity CUPE 3913

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.

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

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

(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.

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.