Question

the refractive index of double convex lens is 1.5 and ratio of radius of curvature is 1:2. the ray coming from point object line is 6cm in front of lens becomes parallel after refraction. find the radius of curvature of its surface.​

Answer 1

Explanation:

From Lens maker's formula,

f

1

=(μ−1)(

R

1

1

−

R

2

1

)

r

1

=20cm,R

2

=−R

1

=−20cm

∴

f

1

=(1.5−1)(

20

1

+

20

1

)

=0.5(

10

1

) or f=20cm

Answer 2

Answer :

• Lens used = Convex
• Refractive index = 1.5
• Ratio of the radius of their curvature = 1:2
• Focal Length = 6 cm
• Radius of curvature = ?

$\quad$━━━━━━━━━━━━━━━━━━━━━━━━━

☯ R₁ : R₂ = 1:2

☯ R₁ = R & R₂ = -2R

$\displaystyle\underline{\bigstar\:\textsf{According to the Question :}}$

$\sf\implies \dfrac{1}{f} = (\mu - 1)(\dfrac{1}{R_1} - \dfrac{1}{R_2})$

$\sf\implies \dfrac{1}{6} = (1.5 - 1)(\dfrac{1}{R} - \dfrac{1}{-2R})$

$\sf\implies \dfrac{1}{6} = (0.5)(\dfrac{1}{R} \times \dfrac{2}{2} + \dfrac{1}{2R})$

$\sf\implies \dfrac{1}{6} = (0.5)(\dfrac{2}{2R} + \dfrac{1}{2R})$

$\sf\implies \dfrac{1}{6} = (0.5)(\dfrac{2+1}{2R})$

$\sf\implies \dfrac{1}{6} = (0.5)(\dfrac{3}{2R})$

$\sf\implies \dfrac{1}{6} = \dfrac{0.5\times 3}{2R}$

$\sf\implies \dfrac{1}{6} = \dfrac{1.5}{2R}$

$\sf\implies 2R = 6\times 1.5$

$\sf\implies R =\dfrac{9}{2}$

$\implies\underline{\boxed{\pink{\mathfrak{r = 4.5 \ cm}}}}$

$\quad$━━━━━━━━━━━━━━━━━━━━━━━━━

Therefore,

• R₁ = 4.5 cm
• R₂ = 4.5 × 2 = 9 cm