Question

vi) An object 4 cm high is placed at a distance of 15 cm in front of a
concave mirror having radius of curvature 10 cm. Where will image
form ?​

Answer 1

GIVEN:

• Height of object, h = 4 cm
• Object distance, u = - 15 cm
• Rasius of curvature of concave mirror , R = - 10 cm

TO FIND:

Image distance, v

FORMULA:

• According to mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

• Focal length = Radius of curvature / 2

SOLUTION:

STEP 1: TO FIND FOCAL LENGTH

$f = \frac{R}{2} \\ \\ \implies f = \frac{ - 10}{2} \\ \\ \implies f = - 5 \: cm$

The focal length of concave mirror, f = -5 cm

STEP 2: TO FIND IMAGE DISTANCE

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ \frac{1}{ - 5} = \frac{1}{v} + \frac{1}{ - 15} \\ \\ - \frac{1}{5} = \frac{1}{v} - \frac{1}{15} \\ \\ - \frac{1}{5} + \frac{1}{15} = \frac{1}{v} \\ \\ \frac{1}{v} = \frac{ - 15 + 5}{15 \times 5} \\ \\ \frac{1}{v} = \frac{ - 10}{15 \times 5} \\ \\ v = - \frac{15 \times 5}{10} \\ \\ v = - \frac{15}{2} \\ \\ v = - 7.5 \: cm$

ANSWER:

Image distance, v = - 7.5 cm.

Image is formed at a distance of 7.5 cm to the left of the concave mirror.

POINTS WE INFER:

• In this case, the object and the image are at same side. Both are at left to the concave mirror.

• Since the incident rays actually meet, they form real image.