91影视

An optical lens is a transparent transmissive optical component that is used to either converge or diverge the light emitting from a object. These transmitted light forms an image of that object.

Thin Lens formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Here,

u = Object distance from lens

v = Image distance from lens

f = Focal length of the lens

Sign Convention:

1. Object Distance (u): (+) in front of lens; (-) in the back of lens
2. Image Distance (v): (-) in front of lens; (+) in the back of lens
3. Focal Length (f): (+) for convergent (convex) lens; (-) for divergent (concave) lens

Lateral magnification (m): It is defined as the ratio of the height of image to the height of the object viewed through a lens. It is a dimensionless quantity.

$m=-\frac{v}{u}=\frac{H_i}{H_0}$

Here,

H_o = Height of the object, and H_i = Height of the image

Sign Convention:

1. If m is positive 鈥� Image is erected
2. If m is negative 鈥� Image is inverted

Focal length of 1^st lens, f₁ = +8.5 cm

Object distance, u = +390 cm

By using lens formula

$$\begin{gathered} 鈬�\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ 鈬�\frac{1}{+390}+\frac{1}{v}=\frac{1}{+8.5} \\ 鈬�\frac{1}{v}=\frac{1}{8.5}-\frac{1}{390} \\ 鈬�\frac{1}{v}=\frac{390-8.5}{390*8.5} \\ 鈬�v=+8.69cm \end{gathered}$$

Therefore, the sensor is +8.69 cm from the lens.

Lateral magnification for a lens is given by

$m=-\frac{v}{u}=\frac{H_i}{H_o}$

Thus, height of the image is given by

$$\begin{gathered} H_i=\left(-\frac{v}{u}\right)H_o \\ =\left(\frac{+8.69cm}{+390cm}\right)*175cm \\ =-3.9cm \end{gathered}$$

Therefore, the image formed is 39 mm in size which will not fit on a screen of 36 mm size.