Question

The semivertical angle of cone of the rays incident on the objective of microscope is 20°. If the wavelength of incident light ray is 6600 $\overset{\circ}{A}$, calculate the smallest distance between two points which can be just resolved.
(Ans : for luminous 10720 $\overset{\circ}{A}$ for nonluminous 8772 $\overset{\circ}{A}$)

Answer 1

heya....

Here is your answer...

Let rr be the radius ll be the slant height
and hh be the vertical height of a cone of semi-vertical angle αα

Surface area S=πrl+πr2S=πrl+πr2------(1)

l=S−πr2πrl=S−πr2πr

The volume of the cone V=13V=13πr2hπr2h

=13=13πr2l2−r2−−−−−√πr2l2−r2

=πr23=πr23(S−πr2)2π2r2−r2−−−−−−−−−−−−√(S−πr2)2π2r2−r2

=πr23=πr23(S−πr2)2−π2r4π2r2−−−−−−−−−−−√(S−πr2)2−π2r4π2r2

=πr23=πr23S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−√πrS2−2πSr2+π2r4−π2r4πr

=r3=r3S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−−−−√S2−2πSr2+π2r4−π2r4

=r3=r3S(S−2πr2)−−−−−−−−−−√S(S−2πr2)

Step 2:

V2=r29V2=r29S(S−2πr2)S(S−2πr2)

V2=S9V2=S9(Sr2−2πr4)(Sr2−2πr4)

dV2dr=S9dV2dr=S9[2Sr−8πr3][2Sr−8πr3]

d2V2dr2=S9d2V2dr2=S9[2S−24πr2][2S−24πr2]------(2)

Now dV2drdV2dr=0=0

⇒S9⇒S9(2Sr−8πr3)=0(2Sr−8πr3)=0

⇒(S−4πr2)=0⇒(S−4πr2)=0

Putting S=4πr2S=4πr2 in (2)

d2Vdr2=S9d2Vdr2=S9[8πr2−24πr2]=−ve[8πr2−24πr2]=−ve

⇒V⇒V is maximum when S=4πr2S=4πr2

Step 3:

Putting the value in equ(1)

4πr2=πrl+πr24πr2=πrl+πr2

4πr2−πr2=πrl4πr2−πr2=πrl

3πr2=πrl3πr2=πrl

3r2=rl3r2=rl

rl=13rl=13

sinα=13sin⁡α=13

l=sin−1(13)l=sin−1⁡(13)

Thus VV is maximum when S=S=constant

α=sin−113

It may help you...☺☺