Light from a burning match propagates left to right, through a thin lens of foca

Question

Light from a burning match propagates left to right, through a thin lens of focal and then thin they are fired 383 m apart. A real image of the flame is the second lens at a distance of 248 cm from the lens. Part (a) How far from the second lead in centimeters, is its optical object located? value expected and expression. Numeric as Part (b) How far is the burning match from the first lens, in centimeters? Numeric A numeric value is expected and not an expression. in a LASIK vision correction, the power of a patient's eye is increased by 2.5D 2.5 D near the point of 25 cm, what was patient's rear point before the procedure in meters? Assuming this produces normal vision (with a the close the depth of the eye is 2 cm. Numeric A numeric value is expected and not an expression.

Explanation / Answer

7.

suppose flame is at distance a from first lens,

Object distance, p1 = a

f1 = 4.4 cm

Applying 1/f1 = 1/p1 + 1/q1

1/4.4 = 1/a + 1/q1

q1 = 4.4a / (a - 4.4)

now this image will work as object for second lens,

p2 = 30.3 - q1 = (25.9a - 133.2)/(a - 4.4)

f2 = 9.9 cm

image distanc, q2 = 24.8 cm

1/f2 = 1/p2 + 1/q2

1/9.9 = (a - 4.4)/(25.9a - 133.2) + 1/24.8

a = 6.44 cm

(A) distance from second lens = 30.3 + 6.44 = 36.7 cm

(B) distance from first lens = 6.44 cm

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