the radius of curvature of a convex lens are 15cm. and 10 cm respectiv

1.	the radius of curvature of a convex lens are 15cm. and 10 cm respectively. if refractive index of lens is 1.5,then find its focal length
Answer» Given : ➳ Radius of curvature of a convex lens are 15cm and 10cm. ➳ RI of lens = 1.5 To Find : ➠ Focal LENGTH of convex lens. Solution : ⇒ Lens maker's formula : $\bigstar\:\underline{\boxed{\bf{\dfrac{1}{<klux>F</klux>}=(<klux>N</klux>-1){\huge(}\dfrac{1}{R_1}-\dfrac{1}{R_2}{\huge)}}}}$ ፨ where, ✏ f DENOTES focal length ✏ n denotes RI of lens ✏ R¹ and R² denotes radii of curvatures We know that, ➛ Radius of curvature of convex mirror is taken positive. ➛ Radius of curvature of concave mirror is taken NEGATIVE. $\longrightarrow\tt\:\dfrac{1}{f}=(n-1){\huge(}\dfrac{1}{R_1}-\dfrac{1}{R_2}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=(1.5-1){\huge(}\dfrac{1}{15}-\dfrac{1}{(-10)}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=0.5{\huge(}\dfrac{2+3}{30}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=0.5\times \dfrac{5}{30}\\ \\ \longrightarrow\tt\:f=\dfrac{60}{5}\\ \\ \longrightarrow\underline{\boxed{\bf{f=+12\:cm}}}$

the radius of curvature of a convex lens are 15cm. and 10 cm respectively. if refractive index of lens is 1.5,then find its focal length

Answer»

Given :

➳ Radius of curvature of a convex lens are 15cm and 10cm.

➳ RI of lens = 1.5

To Find :

➠ Focal LENGTH of convex lens.

Solution :

⇒ Lens maker's formula :

$\bigstar\:\underline{\boxed{\bf{\dfrac{1}{<klux>F</klux>}=(<klux>N</klux>-1){\huge(}\dfrac{1}{R_1}-\dfrac{1}{R_2}{\huge)}}}}$

፨ where,

✏ f DENOTES focal length

✏ n denotes RI of lens

✏ R¹ and R² denotes radii of curvatures

We know that,

➛ Radius of curvature of convex mirror is taken positive.

➛ Radius of curvature of concave mirror is taken NEGATIVE.

$\longrightarrow\tt\:\dfrac{1}{f}=(n-1){\huge(}\dfrac{1}{R_1}-\dfrac{1}{R_2}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=(1.5-1){\huge(}\dfrac{1}{15}-\dfrac{1}{(-10)}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=0.5{\huge(}\dfrac{2+3}{30}{\huge)}\\ \\ \longrightarrow\tt\:\dfrac{1}{f}=0.5\times \dfrac{5}{30}\\ \\ \longrightarrow\tt\:f=\dfrac{60}{5}\\ \\ \longrightarrow\underline{\boxed{\bf{f=+12\:cm}}}$