1-A single bi-concave lens (a lens with two concavesurfaces) made of diamond (in
ID: 1678079 • Letter: 1
Question
1-A single bi-concave lens (a lens with two concavesurfaces) made of diamond (index of refraction n = 2.42) hassurfaces with radii of curvature r1 = -28.0 cm andr2 = 28.0 cm. What is the focal length of the lensin air?2-If an object is placed at p = 17.0 cm from the lens,where is the image? (Use a positive sign for a real image or aminus sign for a virtual image.
3-If the object has a height of h = 2.70 cm, how largeis the image? (Use a positive sign for an upright image or a minussign for an inverted image.)
Explanation / Answer
Given index of refraction n = 2.42
R1= -28 cm
R2 = 28 cm
We know from lens maker formula ( 1/ f) = ( n-1 ) [ ( 1/R2)– ( 1/ R1 ) ]
= 1.42 [ 0.0714 ]
= 0.1014
Focal length f = 9.859 cm
(b). Object distance u = 17 cm
From the relation ( 1/ v ) – ( 1/ u ) = ( 1/ f)
(1/ v ) = ( 1/ f) + ( 1/ u )
= 0.16
Image distance v = 6.24 cm
(3) height of object O = 2.7 cm
Height of image I = ( v / u ) * O
= 0.991 cm
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