Question

A bi-convex lens is formed with two the plano-convex lenses as shown in the figure. Refractive
index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this
bi-convex lens, for an object distance of 40 cm,
the image distance will be​

Answer 1

Question:-

1) A bi-convex lens is formed with two thin piano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this  bi-convex lens, for an object distance of 40 cm, the image distance will be​

To find:-

• The image distance

Formula applied,

• Lens formula,

   $\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Solution:-

$\sf{\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}}$

$\sf{\dfrac{1}{f_1}=(\dfrac{u_l}{u_s}-1)(\dfrac{1}{R_1}\dfrac{-1}{R_2})$

$\sf{\dfrac{1}{f}=(\dfrac{1.5}{1})-1 (\dfrac{1}{14}\dfrac{1}{ \infty})$

$\sf{\dfrac{1}{f}=\dfrac{0.5}{14}=\dfrac{1}{28}} \: \: \: \: (f_1 =28)$

$\sf{\dfrac{1}{f_2}=(\dfrac{1.2}{1}-1) (\dfrac{1}{\infty}-\dfrac{1}{(-14)}]}$

$\sf{\dfrac{1}{f_2}=\dfrac{2}{10} \: [\dfrac{1}{14}]=\dfrac{1}{70} \: \; \: \: f_2=70}$

 Focal length of combination

$\sf{\dfrac{1}{feq}=\dfrac{1}{28}+\dfrac{1}{70}$

             ⇒feq = 20 cm

Lens formula:-

$\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\sf{\dfrac{1}{20}=\dfrac{1}{v}+(\dfrac{1}{-40})$

$\sf{\dfrac{1}{20}-\dfrac{1}{40}=\dfrac{1}{v}$

$\sf{\dfrac{2-1}{40}=\dfrac{1}{v}$

Therefore, V = 40 cm

Answer 2

Answer:

f = 20cm

v = 40cm

Explanation:

$\frac{1}{feq}$ = $\frac{1}{f1}$ + $\frac{1}{f2}$

∴$\frac{1}{f1}$ =[ $\frac{ul}{us}$ - 1 ] [$\frac{1}{R1}$ - $\frac{1}{R2}$]

∴$\frac{1}{f1}$ = [$\frac{1.5}{1}$ - 1 ] [$\frac{1}{14}$ - 1/∞]

∴$\frac{1}{f1}$ = $\frac{0.5}{14}$

∴$\frac{1}{f1}$ = $\frac{1}{28}$

⇒f1 = 28

∴$\frac{1}{f2}$ = [$\frac{1.2}{1}$ - 1 ] [ 1/∞ - $\frac{1}{-14}$]

∴$\frac{1}{f2}$ = $\frac{0.2}{10}$  [ $\frac{1}{14}$ ]

∴$\frac{1}{f2}$ = $\frac{1}{70}$

⇒f2 = 70

Focal length of the combination will be

∴$\frac{1}{feq}$ = $\frac{1}{28}$ - $\frac{1}{70}$

⇒feq = 20cm

Lens formula,

⇒$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

∴$\frac{1}{20}$ = $\frac{1}{v}$ - ( $\frac{1}{-40}$)

∴$\frac{1}{20}$ - $\frac{1}{40}$ = $\frac{1}{v}$

∴$\frac{1}{v}$ = $\frac{1}{40}$

⇒v = 40cm