Bookmark An insect 5.75 mm tall is placed 25.0 cm to the left of a thin planocon

Question

Bookmark An insect 5.75 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 12.9 cm , and the index of refraction of the lens material is 1.70.

a) Calculate the location of the image this lens forms of the insect. the answer is 70.1 cm

b) Calculate the size of the image. the answer 16.1 mm

c)Repeat part A if the lens is reversed. ??

d)Repeat part B if the lens is reversed. ??

i need answers of c and d

Explanation / Answer

Height of object O = 5.75 mm

Distance of the object u = 25 cm

radius of curvature R = 12.9 cm

Focal length 1/f = (n-1) (1/R)

= (1.7-1) (1/12.9)

f = 12.9/0.7

= 18.42 cm

From the relation (1/u)+(1/v) = (1/f)

(1/25) +(1/v) =(1/18.42)

(1/v) = 0.0142

v = 70.1 cm

(b). Magnification m = v / u

= 70.1 / 25

Also we know m = I / O

I / O = 70.1 / 25

I = (70.1/25)(5.75 mm)

= 16.125 mm

(c).

Focal length 1/f = (n-1) (-1/R)

= (1.7-1) (-1/12.9)

f = -12.9/0.7

= -18.42 cm

From the relation (1/u)+(1/v) = (1/f)

(1/25) +(1/v) =(1/-18.42)

(1/v) = -0.0942

v = -10.6 cm

(d). Magnification m = v / u

= -10.6 / 25

Also we know m = I / O

I / O = -10.6 / 25

I = (-10.6/25)(5.75 mm)

= -2.43 mm

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.