IIT JEE , AIEEE Forum - Physics , Optics

Mr.IITIAN007

A converging lens of focal length f_conand a diverging mirror of focal length f_divare placed at a seperation of d.Then what is the focal length of the combination.Note one is lens and the other is mirror.

taruntanuj007

No dont try summming up the powers it wont work here. Here u shud take out images for each case and then proceed basically apply lens and mirror formula repeatedky to get the answer......tooo bad no shortcuts!!!
Cheers!!!

Mr.IITIAN007

I am not stating about the powers here.I am not talking about any shortcut.I know there are many numericals based on 'focal length of the combination' but this is a unique one.I have tried applying the basic formulas but at last two variables are remaining which does not give the correct answer.

taruntanuj007

What variables r remaining u have everything there is no irect formula!!!

akku

Hi IITIAN007 its a ques based on a very simple concept :Rays emerging from an object at the focus of the combination would appear to diverge from the radius of curvature of the diverging mirror.

Hence applying 1/v-1/u=1/f for the lens

1/(2fdiv+d) -1/u=1/fcon

Since the image will b formed to the right of the diverging mirror at R=2fdiv from the mirrors pole , wrt to the lens it will b at distance R+d

=> on solving u=fcon*(2fdiv+d)/(fcon-(2fdiv+d))

Pls tell me if its correct

ITS MY 100TH POST

DO RATE ME!

elessar_iitkgp

If d>f_con then parallel rays converge due to the lens at a point before they reach the mirror. In this case the focal length of the system is f_con .

If d<f_con
For parallel rays striking the lens they converge at focus. Hence for the mirror,
u = -(f_con- d) (A virtual object)
Applying the mirror formula,
1/v + 1/(f_con- d) = -1/f_div
v = - [(f_con- d) f_div ] /(f_con+ f_div-d)
Distance from the optic centre of the lens
[(f_con- d) f_div ] /(f_con+ f_div-d) + d
= (f_con f_div - (f_con+ d)d)/(f_con+ f_div-d)

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