The figure shows an arrangement of an equiconvex lens of refractive index = 1.5and a concave mirror.?

Question

The figure shows an arrangement of an equiconvex lens of refractive index = 1.5and a concave mirror. A point object O is placed on the principle axis at a distance 40 cm from the lens such that the final image is also formed at the position of the object . If the radius of curvature of the concave mirror is 80cm, find the distance d.The focal length of the lens in air is 20 cm.

The surroundings does not have refractive index=1 .....plzz see the diagram for refractive index of the surroundings.....

LINK TO THE IMAGE:.... http://www.goiit.com/posts/list/geometrical-wave-optics-the-figure-shows-an-arrangement-of-an-76008.htm

Answer 1

Let:

p1 be the power of the lens,

u1 be the refractive index of the medium left of the lens,

u2 be the refractive index of the lens material,

u3 be the refractive index of the medium right of the lens,

r1 be the radius of curvature of each face of the lens,

r2 be the radius of curvature of the mirror,

a be the distance from the object to the lens,

b be distance from the lens image to the mirror,

c1, c2, c3, c4 be convergences as described in the calculations.

All convergences and powers are in cm^(-1).

The power of the lens is:

p1 = (u2 - u1) / r1 + (u3 - u2) / (- r1)

= (1.5 - 1.2) / 40 + (2.0 - 1.5) / (- 40)

= - 0.005.

In this situation, the lens is diverging.

The power of the mirror is:

p2 = - 2u3 / (- r2)

= - 2 * 2.0 / (- 80)

= 0.05.

Convergence before the lens is:

c1 = - u1 / a

= - 1.2 / 40

= - 0.03.

Convergence after the lens is:

c2 = c1 + p1

= - 0.03 - 0.005

= - 0.035.

The image is virtual, and left of the lens.

Convergence before the mirror is:

c3 = u3 c2 / (u3 - dc2)

= 2.0 * (- 0.035) / [ 2.0 - d(- 0.035) ]

= - 0.07 / (2.0 + 0.035d).

Convergence after the mirror is:

c4 = c3 + p2

= [ - 0.07 / (2.0 + 0.035d) ] + 0.05.

As the final image coincides with the object, the light rays must trace the same path returning as they do outward.

Equating c3 and c4, but changing the sign of c4 as the diverging light returns converging:

- 0.07 / (2.0 + 0.035d) = [ 0.07 / (2.0 + 0.035d) ] - 0.05

0.14 / (2.0 + 0.035d) = 0.05

d = [ (0.14 / 0.05) - 2.0 ] / 0.035

= 22.857

= 22.9 cm. to 3 sig. fig.