Again, the quantity discharged in the time dt is

V X A X. $ x dt and it is also measured by

- grl a X ds

πνα ds : dt =



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..s= S Xc which being substituted in (1) gives the quantity of air expelled during t", viz. Qs =rS x (a - me

- :), x being a known function of ť. Let x = 0. Then

Qs = aras the whole content, as it ought.






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768. 9,9, denote the distances of the object and its image from the centre of the reflector whose radius is r, we have (see Coddington's Optics, p. 8)


q 9 Also, when the object, 0, is very small, compared with the radius of the reflector, it and its image o' may be considered circular arcs, and in the limit, we have

0:0':: 9:4

ra ::q:

g+ 29

::7 + 29:7. But, if 0, 8' be the angles subtended by 0, 0' at the vertex, we get 0:0 :: (r + 9)0 :(r-9)

ra :: (r+9) 0 : (


pt 29
:: (r + 29)0 ; rd?
::r+29:r:: (r + 29) 0:rl'

..0 =0.


769. If A, A" denote the distances of the object o and image o from the lens; then (Coddington's Optics, p. 120)

0:0 :: A:A" But in the lens (Coddington, p. 61.)


= (m



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F being the distance of the principal focus

FA 0:0 :: A:

:: A+F:F.

A + F Now, by the question, 0 : 0' :: 1:2, and since the image is to be erect, the object and image are on the same side of the lens, and the distance of the principal focus on the other side is negative

:: 1:2:: F-A:F


rtr x

rr" When r =g

A = (m - 1)r.





The principal focal length, measured from the centre is (Coddington, p. 66)

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Let a be the distance between the object and the eye, and x the distance of the lens from the object; then (Code dington, p. 120) the Visual angle 8 oC

(F - A)(" + k)

(Coddington, p. 61.)


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1 .. 0 OC

(F - A) k+ A XF and by the question

(F - A)kt A XF = minimum But A = x, and k = a

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Since the angle of deviation increases with the angle (R) of incidence, (see Wood's Optics, p. 46,) the rod will be most bent, when this latter angle is a maximum. Now, the angle of refraction being 90°, is a maximum, and then

sin. O Em x sin. 90 = m :: generally the maximum deviation is

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773. Letr be the radius of the base, and h the height of the cylinder. Then, if Q, o be the angles of incidence and refraction, we have . p


2r But sin. Q =

w(47% + ha)

3 i. sin. p =

2 4r + h) Therefore by the question,

go + x x tan. Q: 2r :: 3 :h

x being the required depth. Hence,


h (77% + 4h)
2r h tan. 27(7r2 + 4h) – 3h

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774. The density of the sun's rays, or the brightness of his image when viewed with a reflector or refractor cc

area of aperture x power : (see Wood, p. 120.)

(Focal Length) Hence, supposing the power the same in both the reflector and lens we have, by the question


F FM 2r, 2nd being the linear apertures, and F, F the focal lengths. But F = 2F (Wood, pp. 20, 95.)

r ..q

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775. Since the object o is very small, it may be considered an arc of a circle concentric with a vertical section of the cylinder. Hence the image oʻ will also be a circular arc concentric with the object, and if q, q' denote their radii, we have

0:0 :: 2:9. But (Wood, Prop. 43, Cor, 1.)

9-f:f:: 9:4

f being the focal distance ::0:0::9-f:f which gives the magnitude of the image, when q and f are known.

776. In the sphere the angle of incidence at the second surface is always equal to the angle of refraction at the first surface. Hence it will readily appear, upon drawing the figure, that the angle of incidence at the first surface is equal to the angle of refraction at the second, and therefore the ray intercepted between the surfaces will be the base of two opposite isosceles A, whose vertices are at the centre and intersection of the first and last directions of the ray. Now, when the distance of the centre from this radiant chord is given, the chord itself is given, and the

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