Question

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.Part AIs this lens converging or diverging?Part BWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?Express your answer in centimeters to two significant figures or as a fraction.

Answer 1

A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.

Step-by-step explanation:

A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.

Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.

Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:

1. First, we convert the radii of curvature to the appropriate signs as per the lensmaker's convention (positive for convex surfaces when the outside medium is air). R1 = +10cm and R2 = -15cm, since the light exits from the second surface.
2. Next, we plug the values into the equation (1/f) = (nglass - nair) ((1/R1) - (1/R2)) to get the reciprocal of the focal length.

Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:

(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))

(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)

(1/f) = 0.5 * 0.1667cm⁻¹

(1/f) = 0.08335cm⁻¹

Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:

f ≈ 12cm

Answer 2

12 cm

Step-by-step explanation:

We shall use Lens makers formula here which is as follows

`(1)/(F) =(\mu-1) ((1)/(R_1) -(1)/(R_2))`

Put μ = 1.5 , R₁ = 10 cm ,R₂ = - 15 cm ( according to sign convention )

`(1)/(F) =(1.5-1) ((1)/(10) -(1)/(-15))`

= .5 x ( 15 + 10 ) / 15 x 10

=
`(25)/(2*10*15)`

F = 12 cm