Question

A convex lens of radii of curvature 20 cm and 30 cm
respectively. It is silvered at the surface which has
smaller radius of curvature. Then it will behave as
(ug = 1.5)​

Answer 1

AnswEr :

Concave mirror with equivalent focal length -60/11 cm .

Step-by-step explanation :

GivEn :

• μg = 1.5

• Radius of curvature (R1) = 20 cm

• Smaller radius of curvature (R2) = - 30 cm

To find :

Equivalent focal length and the tyle of mirror = ?

SoluTion :

To solve this question we need to know the formula and concept regarding to the focal length of lens as well as focal length of the mirror .

Focal length of the lens is given as, (If μg is given)

➜ 1/F = (μg - 1)(1/R1 - 1/R2)

Also,

Focal length of the mirror is given as, (if radius of curvature is given)

➜ F = R / 2

Now, proceeding towards the question,

Focal length of the lens :

⇛ 1/F = (μg - 1)(1/R1 - 1/R2)

⇛ 1/F = (1.5 - 1)(1/20 - 1/-30)

⇛ 1/F = 0.5 × (-30 - 20 /20×-30)

⇛ 1/F = 0.5 × (-50/-600)

⇛ 1/F = 0.5 × (5/60)

⇛ 1/F = 2.5/60

⇛ F = 60/2.5

⇛ F = 24 cm

Now,

Focal length of the mirror :

⇛ R1 /2

⇛ -20 / 2

⇛ -10 cm.

Equivalent focal length :

⇛ 2 × 1/FL + 1/Fm

⇛ -2/24 + 1/-10

⇛ -2(-10) + 24/24 × -10

⇛ 20 + 24/-240

⇛ 44/-240

⇛ 11/-60

⇛ -60/11 D

Hence, this system behaves like a concave mirror of focal length -60/11 D.