a double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each an object of 5 height is placed at a distance of 10 cm from the lens .find the position ,nature and size of the image

u = -10 cm  r =20 cm  so f = 10 cm

so acc. to lens formula  - 1 /u  +  1 /v  =1 /f

  Put the value of u and f

  1 /v  =- 1 /10  -1 / 10

  v  = -5

 Position  of image  =-5 cm

since it is negative  the image is formed on the same side of the lens and is virtual

size =magnified 

  • -17
r1=20cm r2=-20cm using formula : 1/f=(u-1)(1/r1 - 1/r2) f=20cm Now u=-10 and f=20 cm using lens formula : 1/f = 1/v - 1/u v=-20cm =>virtual,erect image magnification = v/u = 2 => height of image = 2x5 =10cm => magnified image
  • -12
Am sure, this will help u

  • 31
Lens maker's formula

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