PHYSICS 242 BLOCK 12 DRILL SET, SPRING \'18 NAME Slide this drill set under Mart

Question

PHYSICS 242 BLOCK 12 DRILL SET, SPRING '18 NAME Slide this drill set under Marteena 308's door any time before 7:50 AM Friday, April 6, or give me it in Marteena 312 by 8:00 AM that from the find its focal length in cm. ONE EQUATION USED SOLUTION ANSWER Z. A real object placed 16.0 cm from a miror produces a virtual image 8.0 cm from the miror. The object height is 2.2 mm. Find the lateral magnification. Then, from your answer, determine whether the and whether it is reduced, is erect or inverted or the same lheightl. (The answers are the same for a thin lens.) SOLUTION Lateral magnification Circle two below erect inverted reduced enlarged same lheightl as object 3. The tocal points of a diverging lens are 25.4 cm from its center. A virtual image is lens Find the object distance in cm. From your answer's sign, determine ONE EQUATION USEDSLOTOm 352?Tromthe answer's sign, determine whether the object is real or virtual. ANSWER Circle one below Real Virtual (i. e., plane) as shown below. Its index of refraction is 1.52. Use our sign conventions to find its focal length in cm You should complete the sketch of the lens's cross section to determine the subscripts and signs.) ONE EQUATION USED ANSWER Wanted:-SOLUTION cm 5.Use two (hor hree) principal rays (see page 1133) to locate and sketch the image arrow formed by the thin lastic lens of this real object arrow in air. Use solid lines for real rays and a real image arow. Use dashed lines or virtual rays and a virtual image arrow. In the air, the lines muss be straight for full credit.

Explanation / Answer

The focal length of a spherical mirror is equal to half of its radius of curvature

for spherical mirror

focal length=radius of curvature/2

=25.4cm

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