Question

plzzz help me with explanation...​

Answer 1

❏ Formula Used :-

$\setlength{\unitlength}{0.2 cm}\begin{picture}(12,4)\thicklines\put(-15,9.5){\line(1,0){33}}\qbezier(1,18)(-10,10)(1,1)\qbezier(1,18)(1,1)(1,1)\qbezier(1,18)(10,10)(1,1)\qbezier(1,18)(1,1)(1,1)\put(0.3,-1){ f_1 }\put(0.3,18.5){ L_1 }\end{picture}$

let, the focus length of the covex lens $L_1$ is $f_1$

$\setlength{\unitlength}{0.15 cm}}\begin{picture}(12,4)\thicklines\put(-20,9.5){\line(1,0){44}}\qbezier(1,18)(-10,10)(1,1)\qbezier(1,18)(1,1)(1,1)\qbezier(1,18)(10,10)(1,1)\qbezier(1,18)(1,1)(1,1)\put(0.1,-1.2){ f_2 }\put(0.1,18.5){ L_2 }\end{picture}$

let, the focus length of the covex lens $L_2$ is $f_2$

Now, If the covex lenses $L_1$ and $flL_2$ , of focus lengths $f_1$ and $f_2$ ,are placed co-axially and if a be the distance between the optic centre of the both lenses ,

Then a combination of lens is formed ,

if the focus length of this combination is F ,

Then ,

$\sf\boxed{\large{\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{a}{f_1f_2}}}$

Now,

if a=0 (i.e, the lenses are placed adjacently)

$\sf\boxed{\large\red{{\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$

❏ Solution:-

✏ Given:-

• $f_1$ = 30 cm.
• $f_1$ = 30 cm.

✏ To Find:-

• Focal length of the combination, = F

✏ Explanation :-

$\sf\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$\sf\frac{1}{F}=\frac{1}{30}+\frac{1}{30}$

$\sf\frac{1}{F}=\frac{1+1}{30}$

$\sf\frac{1}{F}=\frac{\cancel2}{\cancel30}$

$\sf\frac{1}{F}=\frac{1}{15}$

$\sf\boxed{\large\red{{F=15 \:cm}}}$

∴ Focus length of the combination is = 15 cm.