Question

Part A) A bright object and a viewing screen are separated by a distance of 81.5 cm . At what distance(s) from the object should a lens of focal length 17.0 cm be placed between the object and the screen in order to produce a crisp image on the screen? d=____? cm

Part B) An object is placed 18 cm from a certain mirror. The image is half the size of the object, inverted, and real. What is the radius of curvature of the mirror? Follow the sign conventions.

Answer 1

Part-A : Given that -

focal length of lens, f = 17 cm

bright object & viewing screen separated by a distance, (d₀ + d_i) = 81.5 cm

using a len's formula,

1 / d₀ + 1 / d_i = 1 / f                         

f = (d₀ d_i) / (d₀ + d_i)    { eq.1 }

inserting the values in eq.1,

(17 cm) = (d₀ d_i) / (81.5 cm)

(d₀ d_i) = 1385.5 cm                                         { eq.2 }

where, d_i = (81.5 cm - d₀)

inserting the value of 'd_i' in eq.2,

d₀ [(81.5 cm - d₀)] = 1385.5 cm      

81.5 d₀ - d₀² = 1385.5

rearranging an above eq.

d₀² - 81.5 d₀ + 1385.5 = 0                                    { quadratic equation }

To find the value of d₀ :

nearest distance from the object is given as, d₀ = 24.1 cm

farthest distance from the object is given as, d₀ = 57.3 cm

Part-B : Using a mirror formula -

1 / d₀ + 1 / d_i = 1 / f                                          { eq.3 }

and magnification of the mirror is given as, m = - d_i / d₀

h_i / h₀ = - d_i / d₀ { eq.4 }

we know that the image is half the size of the object, inverted image which is negative.

where, h_i = - h₀ / 2

inserting the value of 'h_i' in eq.3,

(- h₀ / 2) / h₀ = - d_i / d₀

(1/2) = d_i / d₀

d_i = d₀ / 2

where, object distance d₀ = 18 cm

then, d_i = (18 cm) / 2 = 9 cm

image is in positive, which means that image is in front of the mirror (real).

Now, inserting all these values in eq.3,

1 / (18 cm) + 1 / (9 cm) = 1 / f

1 / f = (3 / 18) cm

f = 6 cm

the radius of curvature of the mirror which will be given as :

R = 2 f                                               { eq.5 }

inserting the value of 'f' in eq.5,

R = 2 (6 cm)

R = 12 cm

If focal is in positive which means that the mirror is "concave mirror".