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C PHYS1252, Exam#4A, Solution Steps and Grading Key for Part VI: Ray Diagram (10P) 1P: Draw optical axis, lens principal plane, H, and correctly placed object as arrow to the right of the lens, with all 3 elements properly labeled. 1P: Correctly identify and label incoming side of lens, to the right of lens. Reasoning: i) object is real by sign convention, object is located on incoming side ii) object is placed to the right of lens to the right of lens is the incoming side 1P: Correctly identify and label outgoing side, to the left of the lens: Reasoning: i) The device is a lens with refracted rays leaving opposite to side they entered. ii) Outgoing side is opposite to incoming side: to the left of lens. 1P: Correctly place focal points F and F’: F to the right of the lens and right of object, on the optical axis; F’ at same distance to the left of lens. Label F and F’. Reasoning: i) lens is convergent, i.e., has negative focal length f<0. ii) by sign convention, F is on “in” and F’ is on “out” side, i.e., F is to the right and F’ to the left of lens. iii) F and F’ are at both at a distance |f| from lens F and F’ are at same distance from, and on opposite sides of the lens. iv) absolute object distance from lens, |d|, is less than absolute focal length, |f| v) F is located to the right of the object, i.e., the real object is between F and lens. 2P: Correctly draw one of the three principal ray pairs (see drawing): Either: F-‐F’-‐ray Or: P-‐P’-‐ray Or: C-‐C’-‐ray 2P: Correctly draw a second principal ray pair (see drawing): Either: P-‐P’-‐ray Or: C-‐C’-‐ray Or: F-‐F’-‐ray Note: For a mirror (unlike a lens!) the C-‐ray and C’-‐ray would be two distinct lines, each intersecting the optical axis at the same angle, but one from above, the other from below. So, the C’-‐ray for a mirror would have to be constructed from the C-‐ray, by reflecting the C-‐ray at the optical axis -‐-‐ unlike a lens, where the C-‐ray is simply continued straight through the lens to get the C’-‐ray (see drawing). 1P: Correctly draw image at intersection of two outgoing principal rays as vertical arrow to the right of lens, arrow orientation same as object arrow, i.e., erect relative to object (see drawing). 1P: State: image is virtual. Reasoning: i) image is formed to the right of the lens image is not on the outgoing side ii) by sign convention: image is virtual and d’<0 Note: a virtual image is formed whenever the outgoing rays, (F’, P’, C’) do not physically pass through the image point at which they intersect. Rather, their intersection is obtained by extrapolating the outgoing rays “backwards” to the “not outgoing” side of the device, which then defines the virtual image point. PHYS1252, Exam#4B, Solution Steps and Grading Key for Part VI: Ray Diagram (10P) 1P: Draw optical axis, lens principal plane, H, and correctly placed object as arrow to the left of the lens, with all 3 elements properly labeled. 1P: Correctly identify and label incoming side of lens, to the right of lens. Reasoning: i) object is virtual by sign convention, object is located not on incoming side ii) object is to the left of lens to the left of lens is not the incoming side iii) incoming side is to the right of the lens 1P: Correctly identify and label outgoing side, to the left of the lens: Reasoning: i) The device is a lens with refracted rays leaving opposite to side they entered. ii) Outgoing side is opposite to incoming side: to the left of lens. 1P: Correctly place focal points F and F’: F to the left of the lens and left of object, on the optical axis; F’ at same distance to the right of lens. Label F and F’. Reasoning: i) lens is divergent, i.e., has negative focal length f<0. ii) by sign convention, F is not on “in” and F’ is not on “out” side, i.e., F is to the left and F’ to the right of lens. iii) F and F’ are at both at a distance |f| from lens F and F’ are at same distance from, and on opposite sides of the lens. iv) absolute object distance from lens, |d|, is less than absolute focal length, |f| v) F is located to the left of the object, i.e., the virtual object is between F and lens. 2P: Correctly draw one of the three principal ray pairs (see drawing): Either: F-‐F’-‐ray Or: P-‐P’-‐ray Or: C-‐C’-‐ray 2P: Correctly draw a second principal ray pair (see drawing): Either: P-‐P’-‐ray Or: C-‐C’-‐ray Or: F-‐F’-‐ray Note: For a mirror (unlike a lens!) the C-‐ray and C’-‐ray would be two distinct lines, each intersecting the optical axis at the same angle, but one from above, the other from below. So, the C’-‐ray for a mirror would have to be constructed from the C-‐ray, by reflecting the C-‐ray at the optical axis -‐-‐ unlike a lens, where the C-‐ray is simply continued straight through the lens to get the C’-‐ray (see drawing). 1P: Correctly draw image at intersection of two outgoing principal rays as arrow to the left of lens, arrow oriented opposite to object arrow, i.e., inverted relative to object (see drawing). Note: Depending on choice of object position, the image may be (i) closer the lens than the object, i.e., |d’| < |d|; or (ii) further away, i.e., |d’| > |d|. Case (i) arises if the object position is chosen such that |d|>2|f|; case (ii) arises if |d|<2|f|. 1P: State: image is real. Reasoning: i) image is formed to the left of the lens image is on the outgoing side ii) by sign convention: image is real and d’>0 PHYS1252, Exam#4C, Solution Steps and Grading Key for Part VI: Ray Diagram (10P) 1P: Draw optical axis, lens principal plane, H, and correctly placed object as arrow to the left of the lens, with all 3 elements properly labeled. 1P: Correctly identify and label incoming side of lens, to the right of lens. Reasoning: i) object is virtual by sign convention, object is located not on incoming side ii) object is to the right of lens to the right of lens is not the incoming side iii) incoming side is to the left of the lens 1P: Correctly identify and label outgoing side, to the right of the lens: Reasoning: i) The device is a lens with refracted rays leaving opposite to side they entered. ii) Outgoing side is opposite to incoming side: to the right of lens. 1P: Correctly place focal points F and F’: F to the right of the lens and left of object, on the optical axis; F’ at same distance to the left of lens. Label F and F’. Reasoning: i) lens is divergent, i.e., has negative focal length f<0. ii) by sign convention, F is not on “in” and F’ is not on “out” side, i.e., F is to the right and F’ to the left of lens. iii) F and F’ are at both at a distance |f| from lens F and F’ are at same distance from, and on opposite sides of the lens. iv) absolute object distance from lens, |d|, is greater than absolute focal length, |f| v) F is located to the left of the object, i.e., F is between the virtual object and lens. 2P: Correctly draw one of the three principal ray pairs (see drawing): Either: F-‐F’-‐ray Or: P-‐P’-‐ray Or: C-‐C’-‐ray 2P: Correctly draw a second principal ray pair (see drawing): Either: P-‐P’-‐ray Or: C-‐C’-‐ray Or: F-‐F’-‐ray Note: For a mirror (unlike a lens!) the C-‐ray and C’-‐ray would be two distinct lines, each intersecting the optical axis at the same angle, but one from above, the other from below. So, the C’-‐ray for a mirror would have to be constructed from the C-‐ray, by reflecting the C-‐ray at the optical axis -‐-‐ unlike a lens, where the C-‐ray is simply continued straight through the lens to get the C’-‐ray (see drawing). 1P: Correctly draw image at intersection of two outgoing principal rays as arrow to the left of lens, arrow oriented opposite to object arrow, i.e., inverted relative to object (see drawing). 1P: State: image is virtual. Reasoning: i) image is formed to the left of the lens image is not on the outgoing side ii) by sign convention: image is virtual and d’<0
