An object is placed at x = 0. A converging lens with f1 = 35.0 cm is placed at x

Question

An object is placed at x = 0. A converging lens with f1 = 35.0 cm is placed at x = 60.0 cm and a concave mirror with radius of curvature R = 48.0 cm is placed at x = 120 cm. Considering light that leaves the object, passes through the lens, reflects from the mirror and then passes back through the lens, determine:

(a) the x-coordinate of the final image;

(b) the overall magnification of the final image;

(c) whether the final image is real or virtual;

(d) whether the final image is inverted or upright.

A drawing would be very helpful :) ANSWERS:-69.2 cm, 1.88, real, upright

Explanation / Answer

the object distance u=60 cm

focal length of lens 1=>f1=35 cm

focal length of lens 2=>f2=24 cm

the distance b/w two lenses d=120-60=60 cm

this concept belongs to rayoptices in light

first we find the image formed b/w two lenses

1/v1-1/u=1/f1

1/v1-1/60=1/35

1/v1=1/60+1/35

v1=35*60/60+35 =2100/95=22.1 cm

the image formed b/w the lenses=v1=22.9 cm

now the object distance for second lenses u2=d-v1=60-22.9=37.1 cm

now we find the image formed in second lenses

1/v2=1/u2-1/f2=1/37,8-1/24.

v2=912/13.1.=-69.4 cm

the finial image is formed =-69.4 cm

(b) ans

the over all magnifaction of lens >m=69.4/37.1=1.87

(c) ans

the finial image formed is real because the magnifaction is positive

(d) ans

the finial image formed is upright because the magnifaction is negative

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.