Please help for physics question?

Question

Please help for physics question?

1 Answer

Impact of this question

1315 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T00:37:39

I got option (C)

Explanation:

enter image source here

As described in the problem the given optical system may be treated as a combination of three lenses where there are two equi-convex glass lenses of focal lengths $f_{1} = 10 c m and f_{2} = 20 c m$ $f_1=10cm and f_2=20cm$ in contact and one concave water lens formed by inserted water in between the gap of two glass lenses.

Now let the radius of curvature of the equi-convex lens of focal length $f_{1}$ $f_1$ be $R_{1}$ $R_1$ . So by using lens maker formula we can write

$\frac{1}{f_{1}} = (μ_{g} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{1}})$ $1/(f_1) = (mu_g-1)(1/R_1+1/R_1)$

$\Rightarrow \frac{1}{10} = (\frac{3}{2} - 1) \cdot \frac{2}{R_{1}}$ $=>1/10 = (3/2-1)*2/R_1$

$= R_{1} = 10 c m$ $=R_1=10cm$

Similarly if the radius of curvature of the equi-convex lens of focal length $f_{2}$ $f_2$ be $R_{2}$ $R_2$ . then by using lens maker formula we can write

$\frac{1}{f_{2}} = (μ_{g} - 1) (\frac{1}{R_{2}} + \frac{1}{R_{2}})$ $1/(f_2) = (mu_g-1)(1/R_2+1/R_2)$

$\Rightarrow \frac{1}{20} = (\frac{3}{2} - 1) \cdot \frac{2}{R_{2}}$ $=>1/20 = (3/2-1)*2/R_2$

$= R_{2} = 20 c m$ $=R_2=20cm$

So the water lens will have two concave surfaces with radius of curvature $R_{1} and R_{2}$ $R_1andR_2$

And the focal length of water lens $f_{w}$ $f_w$ will be given by

$\frac{1}{f_{w}} = (μ_{w} - 1) (- \frac{1}{R_{1}} - \frac{1}{R_{2}})$ $1/(f_w) = (mu_w-1)(-1/R_1-1/R_2)$

$\Rightarrow \frac{1}{f_{w}} = (\frac{4}{3} - 1) (- \frac{1}{10} - \frac{1}{20})$ $=>1/(f_w) = (4/3-1)(-1/10-1/20)$

$\Rightarrow \frac{1}{f_{w}} = - \frac{1}{3} \cdot \frac{3}{20} = - \frac{1}{20}$ $=>1/(f_w) = -1/3*3/20=-1/20$

So $f_{w} = - 20 c m$ $f_w=-20cm$

Now if the focal length of the combination of these three lenses be $f_{c}$ $f_c$ then

$\frac{1}{f_{c}} = \frac{1}{f_{1}} + \frac{1}{f_{w}} + \frac{1}{f_{2}}$ $1/(f_c)=1/(f_1)+1/(f_w)+1/(f_2)$

$\Rightarrow \frac{1}{f_{c}} = \frac{1}{10} - \frac{1}{20} + \frac{1}{20}$ $=>1/(f_c)=1/10-1/20+1/20$

$\Rightarrow f_{c} = 10 c m$ $=>f_c=10cm$

Let us consider that the combination forms real image of twice in size of an object when object distance is $u_{1}$ $u_1$ . In this case its image distance will be $- 2 u_{1}$ $-2u_1$ ,as magnification is $- 2$ $-2$ for real image here.

So by lens formula we have

$\frac{1}{- 2 u_{1}} - \frac{1}{u_{1}} = \frac{1}{f_{c}}$ $1/(-2u_1)-1/(u_1)=1/(f_c)$

$\Rightarrow - \frac{1 + 2}{2 u_{1}} = \frac{1}{10}$ $=>-(1+2)/(2u_1)=1/10$

$\Rightarrow u_{1} = - 15 c m$ $=>u_1=-15cm$ (here -ve sign denotes the real object distance)

Again if the combination forms virtual image of twice in size of an object when object distance is $u_{2}$ $u_2$ , then its image distance will be $2 u_{2}$ $2u_2$ ,as magnification is $+ 2$ $+2$ for virtual image here.

So by lens formula we have

$\frac{1}{2 u_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{c}}$ $1/(2u_2)-1/(u_2)=1/(f_c)$

$\Rightarrow \frac{1 - 2}{2 u_{2}} = \frac{1}{10}$ $=>(1-2)/(2u_2)=1/10$

$\Rightarrow u_{2} = - 5 c m$ $=>u_2=-5cm$ (here -ve sign denotes the real object distance)

Important Note proposed by renowned and respected teacher known to us as $A08$ $color(magenta)"A08"$ .

"Two shortcuts.

First for equi-convex lens R=f.

Second when a convex of focal length 20cm and a concave lens of focal length -20cm are placed side by side the focal length of combination is ∞. Hence, equivalent focal length of combination of three lenses reduces to the focal length of first lens."

Science

Math

Humanities

... and beyond

Please help for physics question?

1 Answer

Explanation:

Related questions

Impact of this question