therefore

AF: OF: μ: 1,

(Goodwin's Course, Optics, Art. 39, 3rd edit.)

1848.

(A). The image of a straight line formed by a plane refracting surface is a straight line. Why does a straight rod appear bent when partly immersed in water?

(B). What must be its inclination to the horizon when its apparent portions are inclined to each other at the greatest angle?

Let ABQ (fig. 59) be a rod partly immersed in water, Q any point of the rod under water, qBa the direction after refraction of Then to a the ray from Q which falls in the direction QB. first approximation, q will be the image of Q, and qB will be, by (4), the apparent position of the portion BQ of the rod.

Now the angle between the apparent portions of the rod, viz. the QBq, is also the deviation of the ray QB. But the deviation increases as the angle of incidence increases; the angle therefore between the apparent portions of the rod will be greatest when the angle of incidence of the ray QB is greatest, i.e. when it equals the critical angle for water 48° 30', or when the inclination of the rod to the horizon is 41° 30'.

1850.

(A). Point out the distinction between a real and virtual image.

(B). A plane mirror is placed perpendicular to the axis of a concave spherical reflector, nearer to the principal focus than to the face, and between them; rays from a very distant object fall directly upon the spherical mirror: trace the pencil by which an eye will see the image formed after two reflexions at each mirror.

Let 0 (fig. 60) be the centre of the spherical mirror, F the principal focus half-way between 0 and the mirror, a Cb the plane mirror, CF being less than CA, E the position of the

eye.

1

Take Cq CF; then q, will lie on the right of the mirror. The focus conjugate to q, will be at a point q2, such that

2

Again, take Cq, Cq,; join Eq,, meeting the plane mirror in R1, - R12, meeting the spherical mirror in R, - R2g, meeting, when produced, ab in R, FR, meeting, when produced, the spherical mirror in R1; and lastly, draw RQ parallel to 40.

4

Then will QRRRRE be clearly the course of the axis of the pencil by which the eye at E sees the distant object after two reflexions at each of the mirrors.

The image seen will be at 9, which, since the rays do not actually pass through it, is a virtual one.

1850.

(A). A ray of light is refracted through a prism in a plane perpendicular to its edge; find the deviation produced by the refraction.

(B). A speck is situated just within a glass sphere; shew how much of the surface of the sphere must be covered in order that the speck may be invisible at all points outside the sphere on a line drawn from the speck through the centre.

If o,, be the angles of incidence and emergence of a ray which passes through a cone whose vertical angle is i, the deviation = $ + y − i.

Let now A (fig. 61) be a speck just within a glass sphere ADB. Let AD be a ray which, after refraction, is parallel to BC; this will clearly be the ray which makes the greatest angle with AB of all those which enter an eye situated in BC. Draw the two tangent planes to the sphere at D and A, meeting in E. Then, (since it is immaterial whether we consider it as just inside or just outside the surface,) ADF may be regarded as the course of a ray refracted through the prism AED; and our object is to find how this prism must be situated in order that the emergent ray may be parallel to BC.

In order that the emergent ray may be parallel to BC, the deviation must = LDAO = 4;

Hence the portion of the sphere which must be covered is that which subtends an angle DOD' = 24D0B = 4 cos1μ.

1851.

-11

(4). Shew how to find by experiment the focal length of a lens.

(B). The least distance between an object and its image formed by a plano-convex lens of glass is 12 inches; the index of refraction being, find the radius of the spherical surface.

The focal length of a convex lens may be found experimentally by observing the least distance between an object in the axis of the lens and its image formed by the lens. For let Q (fig. 62) be the object, q its image, AQ = u, Aq = v, f focal length of lens.

which shews (since u must be a possible quantity) that the least value of x = 4f.

Thus fof least distance between Q and its image. If the lens be concave, the focal length may be determined by the same experiment, by placing it in contact with a convex lens. (See Griffin's Optics, 2nd edit., Art. 167).

In (B) we have given that the least distance between an object and its image is 12 inches;

therefore

f=23 inches.

4

But if r be the radius of the curved surface,

1848.

(A). Describe the Astronomical Telescope; trace the course of a pencil of rays from any point of a distant object, and find the magnifying power.

(B). If the focal lengths of the lenses be 12 inches and 1 inch, how far must the eye-glass be moved for viewing an object at a distance of 40 feet from the object-glass?

Let A, a (fig. 63) be the object and eye glasses of an Astronomical Telescope, p the image of an object 40 feet distant from A.

Then, since A is a convex lens of 12 inches focal length,

And since ap 1 inch; therefore Aa 13+ inches.

=

1/3

4 18

Now when the instrument is in adjustment for viewing a

distant object, Aa = 13 inches. Hence the eye-glass has to be

pulled out of an inch.

4 Τ