Question

The focal length of a convex lens is 10 cm and its
refractive index is 1.5. If the radius of curvature of
one surface is 7.5 cm, the radius of curvature of
the second surface will be
(1) 7.5 cm
(2) 15.0 cm (3) 75 cm
(4)5.0 cm


pls do full calculation..​

Answer 1

Answer:

OPTION 2

EXPLANATION :::

★ BY FOCUS FORMULA ;;;

★ 1 / F = ( μ - 1 ) { 1/R¹ - 1/R² }

→F = Focal length of lens

→ R¹ = Radius of Curvature of surface 1

→ R² = Radius of Curvature of surface 2

→ μ = Refractive index of Lens

★ By Formula given above ;

»1 / 10 = ( 1.5 - 1 ) { 1 / 7.5 - 1/ R² }

» 0.1 = 0.5 { 1 / 7.5 - 1/ R² }

» 0.1 / 0.5 = { 1 / 7.5 - 1 /R² }

» 1 / 5 = { 1 / 7.5 - 1 / R² }

★ By taking L.C.M. we get ;

» 5 = { 7.5 × R² / 7.5 + R² }

» 37.5 + 5R² = 7.5 R²

» 2.5 R² = 37.5

«★» R² = 15 cm

»» HENCE ; VALUE OF RADIUS OF CURVATURE OF 2nd Surface is " 15cm ".

Answer 2

Answer hidden:

(2) 15.0 cm

Explanation :

➡️By focus formula :

1/ F= (U-1){1/R1- 1/{ R2/ }

F= focal length of lens .

R= radius of curvature .

U= refractive index lens

By formula given above :

1/10=(1.5-1){1/7.5- 1/R2

0.1=0.5={1/7.5-1/R2}

0.1/0.5={1/7.5-1/R2}

By taking L.C.M

We get:

5=(7.5 × R2/7.5 + R2)

37.5 + 5 R2= 7.5 R2

2.5 R2= 37.5

R2=15 cm

So answer is 15 cm .

Thanks .