mittal (im5936) – reflection refraction imaging HW – yeazell – (58010)902710.0pointsAn ordinary magnifying glass produces animage which isThe magnification isM=-qp=--11.424 cm35.7 cm= 0.32.Theimageisvirtual,uprightand11.424 cm in front of the lens.029(part2of3)10.0pointsFind the image distance for an object distanceof 19.9 cm.
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4.virtual and inverted5.real and erectExplanation:A magnifying glass is a double convex lens.The object is located just inside the focallength. Thus the image is virtual, erect, andthe magnification is larger than 1.028(part1of3)10.0pointsA diverging lens has a focal length of 16.8 cm.2f2fffLocate the image for object distance of35.7 cm .Correct answer:-11.424 cm.Explanation:Let :p= 35.7 cmandf=-16.8 cm.2ffFrom the thin lens equation1p+1q=1fq=fpp-f=-(16.8 cm) (35.7 cm)35.7 cm-(-16.8 cm)=-11.424 cm.Correct answer:-9.10954 cm.Explanation:Let :p= 19.9 cm.q=-(16.8 cm) 19.9 cm19.9 cm-(-16.8 cm)=-9.10954 cm.The magnification isM=-qp=--9.10954 cm19.9 cm= 0.457766.Theimageisvirtual,uprightand9.10954 cm in front of the lens.030(part3of3)10.0pointsFind the image distance for an object distanceof 6.9 cm.Correct answer:-4.89114 cm.Explanation:Let :p= 6.9 cm.q=-(16.8 cm) 6.9 cm6.9 cm-(-16.8 cm)=-4.89114 cm.The magnification isM=-qp=--4.89114 cm6.9 cm= 0.708861.