a rod(glass) has convex surfaces on both ends and n= 1.6. The radius of curvatur

Question

a rod(glass) has convex surfaces on both ends and n= 1.6. The radius of curvature at the left=6.00cm and at the right=12.0cm. the length between them is 40 cm. The object for the left is an arrow that is 23 cm to the left of this surface. The arrow is 1.5mm tall and at right angles to the axis. what consitutes the object for the surface at the right end of the rod, what is the object distance for this surface, is the object for this surface real or virtual, and what is the position of the final image

Explanation / Answer

The refractive index of the material n = 1.6 The radius of curvature on the left is R1 = 6 cm The radius of curvature on the right is R2 = -12 cm The thickness of the lens is d = 40 cm The object distance is u1 = 23 cm The height of the object h = 1.5 mm The object for the right side curvature is the image formed by the left side curvature The equation for the curvture at the left is (n-1)/R1 = 1/u1 + n/v1 (1.6-1)/(6 cm) = 1/(23 cm) + 1.6/v1 v1 = 28.3 cm The image is formed is at 28.3 cm from the vertex of the curvature The image is real and inverted with respect to the object This will be object for the curvature on the right at 40 cm - 28.3 cm = 11.7 cm The equation for the curvature at the left is (1-n)/R2 = n/u2 + 1/v2 (1-1.6)/(12 cm) = 1.6/(11.7 cm) + 1/v2 v2 = -5.35 cm The image will be at 5.35 cm from the right curvature The image is virtual and upright with respect to the first image i.e. inverted and real with respect to the object The radius of curvature on the right is R2 = -12 cm The thickness of the lens is d = 40 cm The object distance is u1 = 23 cm The height of the object h = 1.5 mm The object for the right side curvature is the image formed by the left side curvature The equation for the curvture at the left is (n-1)/R1 = 1/u1 + n/v1 (1.6-1)/(6 cm) = 1/(23 cm) + 1.6/v1 v1 = 28.3 cm The image is formed is at 28.3 cm from the vertex of the curvature The image is real and inverted with respect to the object This will be object for the curvature on the right at 40 cm - 28.3 cm = 11.7 cm The equation for the curvature at the left is (1-n)/R2 = n/u2 + 1/v2 (1-1.6)/(12 cm) = 1.6/(11.7 cm) + 1/v2 v2 = -5.35 cm The image will be at 5.35 cm from the right curvature The image is virtual and upright with respect to the first image i.e. inverted and real with respect to the object The equation for the curvature at the left is (1-n)/R2 = n/u2 + 1/v2 (1-1.6)/(12 cm) = 1.6/(11.7 cm) + 1/v2 v2 = -5.35 cm The image will be at 5.35 cm from the right curvature The image is virtual and upright with respect to the first image i.e. inverted and real with respect to the object

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