General Physics III - Practice Exam II Solutions - Fall ... - Amazon S3

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$1 Ch. 22 â Reflection and Refraction of Light 22.1 The ... - Physics$

4A. Two converging lenses (lens 1 and lens 2) with focal length f 1 = 20.0 cm and f 2 = 25.0 cm are placed20.0 cm apart along their common optical axis. An object is placed 60.0 cm in front of lens 1.Determine:(a) the position of the image from lens 1, if it is real or virtual, upright or inverted and its magnification;(b) the position of the final image from lens 2, if it is real or virtual, upright or inverted and its magnification.(c) Sketch a diagram showing the two lenses, the object (as an upright arrow), the image from lens 1and the final image.Solution:(a) The image from the first lens is:1 1 1 1 1 1q 1f 1p 120cm 60cm 30.0 cm⇒ q = + 30.0 cm.1The image is real since q 1 is + ; the image is inverted since m = - q 1 /p 1 is negativem = - (30.0 cm) / (60.0 cm) = -0.50(b) This image is the object for the second lens.Because it is beyond the second lens, it has a negative object distance:p 2= 20.0 cm – 30.0 cm = – 10.0 cm.We find the image formed by the refraction of the second lens:1 1 1+ =p2 q2 f2[1/(– 10.0 cm)] + (1/q 2) = 1/25.0 cm, which gives q 2= + 7.14 cm.The image is real since q 2 is +The total magnification is the product of the magnifications for the two lenses:m = m1m2 = (–q 1 /p 1 )(– q 2 /p 2 ) = q 1 q 2 / p 1 p 2= (+ 30.0 cm)(+ 7.14 cm)/(+ 60.0 cm)(– 10.0 cm) = – 0.357 ( image inverted since m is -).(c)OI 2 I F 1 2F 1F F21
4B. Violet light of wavelength 415 nm passes through a single narrow slit and creates a central diffraction peakwhich has a width (distance between the first minima on both sides of the central maximum) of 9.20 cm ona screen that is 2.55 m from the slit. How wide is the slit?Solution:tan θ1min = ½ (9.20 cm)/(255 cm) = 0.0180 = sin θ1min , because the angle is small.We find the slit width from:a sin θ1min = mλ;a (0.0180) = (1)(415 × 10 –9 m), which gives a = 2.30 × 10 –5 m =0.0230 mm.4C. How many lines per centimeter must diffraction grating have such that there will be no 2 nd order maximumfor light of any wavelength in the visible region (400 nm to 700 nm)?Solution:Because the angle increases with wavelength, to miss a complete order we use the smallest visiblewavelength. The maximum angle is 90°. We find the slit separation fromd sin θ = mλ;d sin 90° = (2)(400 × 10 –9 m), which gives d = 8.00 × 10 –7 m = 8.00 × 10 –5 cm.The number of lines/cm is1/d = 1/(8.00 × 10 –5 cm) = 12,500 lines/cm.
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Page 3: 3.A. Two students are performing an

General Physics III - Practice Exam II Solutions - Fall ... - Amazon S3

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