The radii of curvature of a biconvex lens are 4 cm and 15 cm. the lens is in air, its index of refraction is 1.5. an object is at 1 m before the front surface of the lens. calculate the distance of the image from the back surface of the lens.

v = 6.315 cm

Explanation:

given,

R₁ = 4 cm = 0.04 m

R₂ = 15 cm = 0.15 m

n =1.5

v = 0.06315 m

v = 6.315 cm  

hence, the distance of the image from the back surface is v = 6.315 cm

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards