I found an answer from spacemath.gsfc.nasa.gov

A Guide to Smartphone Sensors

Oct 30, 2018 **...** apps and examine this data using a variety of **mathematical** operations.
Smartphone apps can ... Experiment **2**: Finding the Smartphone Magnetometer!
... iPhone/iPad, allows you to **monitor** the strength of magnetic **field**. **Displays** ...
Outside **house** with leaves. 70. 75. **35**. 59. Outside **house** no leaves. 76. 73.

For more information, see A Guide to Smartphone Sensors

I found an answer from goprep.co

The photograph of a house occupies an **area** of 1.75 cm^2 on a 35 ...

The slide is projected on **to** a screen, and the area **of the** house on the screen is
... **NCERT** - **Physics** Part-I ... **Units** and **Measurements** ... **Formula**: **Areal**
**magnification** (m) = Aimage / Aobject , ... Since the **dimension** of **areal**
**magnification** is L^{2} and **dimension** of **linear** ... The radius of a cHC Verma -
Concepts of **Physics** Part 1.

For more information, see The photograph of a house occupies an **area** of 1.75 cm^2 on a 35 ...

The ratio of the image's area to the object's area is known as** arial magnification ( m_a ).**

The ratio of the image's distance to the object's distance is known as** linear magnification( m_l .**

Relation between the linear and arial magnification:

The square of arial magnification is equal to linear magnification.

m_l=\sqrt{m_a}

Given that

Area of the house on the screen (image) = 1.55 m^2 = 1.55 * 10^4 cm^2

Photograph of a house Area (object) = 1.75 cm^2

Step 1: Determining the arial magnification

\text{ Arial magnification } = \frac{\text{ image's area }}{\text{ object's area }}

\text{ Arial magnification } = \frac{1.55 * 10^4}{1.75}

\text{ Arial magnification } = 8857

Step 2: To find the linear magnification, recall the relationship between linear and arial magnification.

linear magnification m_l = \sqrt{m_a}

m_l = \sqrt{8857}

m_l = 94.1

Hence, linear magnification of the projector-screen m_l = 94.1